text:
from django.test import TestCase
from tx_highered import models

class PriceTrendsTestCase(TestCase):
    def setUp(self):
        # TODO Replace with Factory Boy
        self.ut = models.Institution.objects.get(name__endswith='at Austin')
        self.price_trends = self.ut.latest_tuition

    def test_in_state_maps_to_field(self):
        self.assertEqual(self.price_trends.in_state,
                self.price_trends.tuition_fees_in_state)

    def test_out_of_state_maps_to_field(self):
        self.assertEqual(self.price_trends.out_of_state,
                self.price_trends.tuition_fees_outof_state)

    def test_a_decade_ago_returns(self):
        a_decade_ago = self.price_trends.a_decade_ago
        self.assertTrue(a_decade_ago.year, self.price_trends.year - 10)

    def test_in_state_has_risen_true(self):
        self.assertTrue(self.price_trends.in_state_has_risen)

    def test_in_state_has_risen_false(self):
        old = self.price_trends.a_decade_ago
        old.tuition_fees_in_state = self.price_trends.in_state + 1
        self.assertFalse(self.price_trends.in_state_has_risen)

    def test_out_of_state_has_risen_true(self):
        self.assertTrue(self.price_trends.out_of_state_has_risen)

    def test_out_of_state_has_risen_false(self):
        old = self.price_trends.a_decade_ago
        old.tuition_fees_outof_state = self.price_trends.out_of_state + 1
        self.assertFalse(self.price_trends.out_of_state_has_risen)

    def test_in_state_change(self):
        old = self.price_trends.a_decade_ago
        old.tuition_fees_in_state = self.price_trends.in_state * 0.5
        self.assertEqual(100, self.price_trends.in_state_change)

        old.tuition_fees_in_state = self.price_trends.in_state * 0.25
        self.assertEqual(300, self.price_trends.in_state_change)

        old.tuition_fees_in_state = self.price_trends.in_state * 1.1
        self.assertEqual(10, self.price_trends.in_state_change)

    def test_out_of_state_change(self):
        old = self.price_trends.a_decade_ago
        old.tuition_fees_outof_state = self.price_trends.out_of_state * 0.5
        self.assertEqual(100, self.price_trends.out_of_state_change)

        old.tuition_fees_outof_state = self.price_trends.out_of_state * 0.25
        self.assertEqual(300, self.price_trends.out_of_state_change)

        old.tuition_fees_outof_state = self.price_trends.out_of_state * 1.1
        self.assertEqual(10, self.price_trends.out_of_state_change)

class TestScoreManagerTestCase(TestCase):
    def setUp(self):
        # TODO Replace with Factory Boy
        self.institute = models.Institution.objects.get(
                name__endswith='at Austin')

    def test_latest_returns_most_recent(self):
        latest = (models.TestScores.objects
                .filter(institution=self.institute).latest('year'))
        self.assertEqual(latest, self.institute.testscores.latest())

    def test_oldest_returns_oldest_model(self):
        oldest = (models.TestScores.objects
                .filter(institution=self.institute).order_by('year')[0])
        self.assertEqual(oldest, self.institute.testscores.oldest())

class AdmissionsManagerTestCase(TestCase):
    def setUp(self):
        # TODO Replace with Factory Boy
        self.institute = models.Institution.objects.get(
                name__endswith='at Austin')

    def test_latest_returns_most_recent(self):
        latest = (models.Admissions.objects
                .filter(institution=self.institute).latest('year'))
        self.assertEqual(latest, self.institute.admissions.latest())

    def test_oldest_returns_oldest_model(self):
        oldest = (models.Admissions.objects
                .filter(institution=self.institute).order_by('year')[0])
        self.assertEqual(oldest, self.institute.admissions.oldest())

class AdmissionsTestCase(TestCase):
    def setUp(self):
        # TODO Replace with Factory Boy
        self.institute = models.Institution.objects.get(
                name__endswith='at Austin')


text:
Q: How to export/push my local database in MongoDB atlas I've got a database in my local machine and I've created a cluster in MongoDB atlas and also connect it. I want to export/push my DB to the cluster.I created the dump of my database using this command mongodump --db all_product --collection product
My database:
    > show dbs
    admin        0.000GB
    all_product  0.000GB
    config       0.000GB
    local        0.000GB
    > use all_product
    switched to db all_product
    > show collections
    product
    > db.product.count()
    522

I tried the binary export mongorestore using this command:
C:\Windows\system32>mongorestore --host allProducts-shard-0/allproducts-shard-00-00-k6ufl.mongodb.net:27017,allproducts-shard-00-01-k6ufl.mongodb.net:27017,allproducts-shard-00-02-k6ufl.mongodb.net:27017 --ssl --username shakib --password <testpass> --authenticationDatabase admin

but I'm getting this error - The system cannot find the file specified
How to set the file path or what should I do. My mongo DB files are stored in C:\data\db

A: You need to specify the name of the directory using --dir explicitly.
Refer the code below
mongorestore --ssl --host <host> --authenticationDatabase admin --dir="<dumpDirectory>" -u <adminUserName> --password <password>

<host>: get from mongo atlas dashboard (click on the cluster name) ex: cluster0-shard-00-00-cbei2.mongodb.net:27017
<dumpDirectory>: The folder where database is dumped using mongodump
<adminUserName>: admin User for that database
<password>: admin associated password

redpajama_set_name: RedPajamaStackExchange


text:
\section{The algebra {\em qA} is Koszul}
\label{sec:appendix}

In this appendix we prove that the quadratic algebra $qA$, defined as
the quotient of $\mathbf{T}(\mathgth{g})$ by the two sided ideal generated by the
linear subspace $qR$ of $\mathgth{g}^{\otimes 2}$ spanned by
$\{hg-gh~|~h\in\mathgth{h},g\in\mathgth{g}\}$, is Koszul.  We refer to \cite{BG} and
references therein for the many definitions of Koszulity and their
equivalence.

The Koszul complex $K(qA)$ of $qA$ is a subcomplex of the Bar resolution $B_{qA}(\mathbf{k})$ of $\mathbf{k}$
as a left $qA$-module via the augmentation map $\varepsilon:qA\to\mathbf{k}$:
$$
K(qA):=\bigoplus_{i\geq0}\left(qA\otimes \tilde{K}^i(qA)\right)[i]
\subset\bigoplus_{i\geq0}\left(qA\otimes qA^{\otimes i}\right)[i]=: B_{qA}(\mathbf{k}),
$$
where
$$
\tilde{K}^i(qA):=\bigcap_{k=0}^{i-2}\mathgth{g}^{\otimes k}\otimes qR\otimes\mathgth{g}^{\otimes i-k-2}.
$$
Recall that the differential on the Bar resolution $B_{qA}(\mathbf{k})$ is defined by
\[ d(a_0\otimes\cdots\otimes a_i)=\sum_{k=0}^{i-1}(-1)^ka_0\otimes\cdots\otimes a_ka_{k+1}\otimes\cdots\otimes a_i+(-1)^ia_0\otimes\cdots a_{i-1}\varepsilon(a_i).
\]
\begin{Proposition}
For any $i<0$, $H^i\big(K(qA)\big)=0$. In other words the algebra $qA$ is Koszul.
\end{Proposition}

\begin{Proof}
For any $i>0$, $\tilde{K}^i(qA)=\big(\wedge^{i-1}\mathgth{h}\big)\wedge\mathgth{g}$ is the image of
$\mathgth{h}^{\otimes(i-1)}\otimes\mathgth{g}$ in $\mathgth{g}^{\otimes i}\subset qA^{\otimes i}$ through the total antisymmetrization map
$$
x_1\otimes\cdots\otimes x_i\mapsto
x_1\wedge\dots\wedge x_i:=\sum_{\sigma\in S_i}\epsilon(\sigma)x_{\sigma(1)}\otimes\cdots\otimes x_{\sigma(i)}.
$$
As usual, $\tilde K^0(qA)=\mathbf{k}$.
Now observe that the only non-zero term in the restriction of the
differential of the Bar resolution to
$K(qA)$ is the first one:
$$
d(\sum_s a_0^{(s)}\otimes\cdots\otimes a_i^{(s)})=
\sum_s
a_0^{(s)}a_1^{(s)}\otimes\cdots\otimes a_i^{(s)}.
$$
This is a general fact that is not specific to the peculiar situation
we are working at.

In other words, for an element $a\otimes x_1\wedge\dots\wedge x_i\in
qA\otimes\tilde{K}^i(qA)$ we have
\[
d(a\otimes x_1\wedge\dots\wedge x_i)
=\sum_{j=1}^i(-1)^{j-1}ax_j\otimes x_1\wedge\dots\wedge\hat{x_j}\wedge\dots\wedge x_i.
\]
Remember that the symmetric algebra $S(\mathgth{h})$ is Koszul: its Koszul complex $K\big(S(\mathgth{h})\big)$, which is $\oplus_{i\geq0}S(\mathgth{h})\otimes\wedge^i(\mathgth{h})[i]$
with differential being given by the formula above, is acyclic in negative degrees.
Finally, one sees that $K(qA)$ is isomorphic to the $dg$ $K\big(S(\mathgth{h})\big)$-module freely generated
by the two step complex
$$
\cdots\longrightarrow0\longrightarrow\big(\mathbf{T}(\mathgth{n})\otimes\mathgth{n}\big)[1]\longrightarrow\mathbf{T}(\mathgth{n})\longrightarrow0\longrightarrow\cdots,
$$
which is acyclic in negative degrees.  This can be easily seen by
considering a $\mathbf{k}$-linear splitting $\mathgth{n}\hookrightarrow\mathgth{g}$
of $\mathgth{g}\twoheadrightarrow\mathgth{n}$ and observing that
\begin{align*}
K\big(S(\mathgth{h})\big)\otimes\big(\mathbf{T}(\mathgth{n})\otimes(\mathgth{n}[1]\oplus\mathbf{k})\big)
& \cong
qA\otimes\left(\bigoplus_{i\geq0}\wedge^i(\mathgth{h})[i]\right)\otimes(\mathgth{n}[1]\oplus\mathbf{k})
\\
& \cong qA\otimes\left(\mathbf{k}\oplus\bigoplus_{i\geq0}\big(\wedge^{i-1}(\mathgth{h})\big)\wedge\mathgth{g}[i]\right).
\end{align*}
We leave to the reader the straightforward task of tracking the
differential through this identification.  \qed
\end{Proof}

\section{An example of a non trivial class}
\label{ex}

We now give an example of an inclusion of Lie algebras
$\mathfrak{h}\hookrightarrow\mathfrak{g}$ for which the obstruction
class is non trivial.  Let $\mathfrak{g}=\mathfrak{sl}_2$; recall that
it is generated by $e$, $h$ and $f$, satisfying the relations
$$
[e,f]=h\,,\quad[h,e]=2e\,,\quad[h,f]=-2f\,.
$$
Now let $\mathfrak{h}$ be the Lie subalgebra generated by $e$ and $h$. Then $\mathfrak{n}=\mathfrak{g}/\mathfrak{h}$ is the one dimensional $\mathfrak{h}$-module generated as a vector space by $f$, with module structure defined by
$$
e\cdot f=0\quad\textrm{and}\quad h\cdot f=-2f\,.
$$
\begin{Proposition}
The obstruction class $\alpha$ is non-trivial.
\end{Proposition}

\begin{Proof}
First observe that the Chevalley-Eilenberg 1-cocycle
\[ c\in C^1\big(\mathfrak{h},\Hom(\mathfrak n,\mathfrak h)\big) \]
given by
$$
c(e)(f)=e \cdot f-[e,f]=-h,\quad c(h)(f)=h\cdot f-[h,f]=0
$$
is a representative of the exact sequence
$$
0\rightarrow \mathfrak h\rightarrow\mathfrak g\rightarrow\mathfrak n\rightarrow 0\,.
$$
Therefore the $1$-cocycle $a\in C^1\big(\mathfrak{h},{\rm Hom}(\mathfrak n^{\otimes2},\mathfrak n)\big)$ given by
$$
a(e)(f,f)=-h\cdot f=2f,\quad a(h)(f,f)=0
$$
is a representative of the obstruction class $\alpha$.

Finally, observe that since $e$ acts trivially on $\mathfrak n$, then it acts trivially on ${\rm Hom}(\mathfrak n^{\otimes2},\mathfrak n)$. Consequently,
for any $b\in{\rm Hom}(\mathfrak n^{\otimes2},\mathfrak n)$ we have
$d(b)(e)=0$, so that $a\neq d(b)$.  Thus $\alpha\neq0$.
\qed
\end{Proof}

\section{PBW for inclusions into first order neighborhoods}
\label{sec:fon}

In this section we study the ${\mathsf{PBW}}$ property for the inclusion
$j:\mathgth{h}\hookrightarrow\mathgth{h}^{(1)}$ of $\mathgth{h}$ into its first order
neighborhood Lie algebra $\mathgth{h}^{(1)}$. We prove that the ${\mathsf{PBW}}$ theorem
holds if and only if the extension class $\alpha$, defined in
Section~\ref{sec:hkr}, vanishes.

\paragraph
We begin with some notation that will be used. Denote the Lie algebra
inclusion $\mathgth{h}\hookrightarrow \mathgth{g}$ by $i$. Denote the natural maps of
Lie algebras $\mathgth{h}\rightarrow \mathgth{h}^{(1)}$ and $\mathgth{h}^{(1)}\rightarrow
\mathgth{g}$ by $j$ and $k$ respectively so that $i=k\circ j$.  Denote by
$i^*$ the restriction functor from $\mathgth{g}$-modules to $\mathgth{h}$-modules and
by $i_!$ the induction functor in the reverse direction. Thus we have
the adjunction $i_!\dashv i^*$. We also have similar functors and
adjunctions for the maps $j$ and $k$.  Finally we denote the
$1$-dimensional trivial representation of the Lie algebra $\mathgth{h}$ by
${\mathbf 1}_\mathgth{h}$.

\paragraph
The goal of the current paper is to understand ${\mathsf{PBW}}$ properties for
\[
i^*i_!({\mathbf 1}_\mathgth{h})=\mathbf{U}(\mathgth{g})\otimes_{\mathbf{U}(\mathgth{h})}{\mathbf 1}_\mathgth{h}=\mathbf{U}(\mathgth{g})/\mathbf{U}(\mathgth{g})\mathgth{h}. \]
In this section we study the object $j^*j_!({\mathbf 1}_\mathgth{h})$ which is
easier to understand. This representation can be described as a quotient
of the tensor algebra $\mathbf{T}(\mathgth{g})$:
\[j^*j_!({\mathbf 1}_\mathgth{h})=\mathbf{U}(\mathgth{h}^{(1)})\otimes_{\mathbf{U}(\mathgth{h})}{\mathbf 1}_\mathgth{h}=\mathbf{T}(\mathgth{g})/\left(J+\mathbf{T}(\mathgth{g})\mathgth{h}\right).\]
Here $J$ denotes the two sided ideal generated by $hg-gh-[h,g]_\mathgth{g}$ for
$h\in\mathgth{h}$ and $g\in\mathgth{g}$.

\paragraph
There are two natural increasing filtrations on the $\mathgth{h}$-module
$j^*j_!({\mathbf 1}_\mathgth{h})$.  The first one is induced from the natural
filtration on $\mathbf{U}(\mathgth{h}^{(1)})$, for which elements of $\mathgth{h}^{(1)}$ have
degree $1$.  The second one is induced by the grading on $\mathbf{T}(\mathgth{g})$,
where elements of $\mathbf{T}^k(\mathgth{g})$ have degree $k$.  Throughout this paper
we shall only work with the latter filtration, which shall be denoted
by $F^0\subset F^1\subset F^2 \cdots \subset F^k \cdots$.  Explicitly,
$F^k$ consists of those elements of $j^*j_!({\mathbf 1}_\mathgth{h})$ that have a
lift to $\mathbf{T}(\mathgth{g})$ of degree $\leq k$.

\begin{Lemma}
\label{lem:2}
The associated graded $\mathgth{h}$-module $\gr(F^\cdot)$ of the above filtration is
precisely $\mathbf{T}(\mathgth{n})$.  In other words the successive quotients
$F^k/F^{k-1}$ are isomorphic, as $\mathgth{h}$-modules, to $\mathgth{n}^{\otimes k} $.
\end{Lemma}

\begin{Proof}
As $j^*j_!{\mathbf 1}_\mathgth{h}$ is a quotient of $\mathbf{T}(\mathgth{g})$ by the sum of two
ideals, we will understand this quotient in two steps corresponding to
the two succesive quotients.

We first give a description of the algebra $A = \mathbf{T}(\mathgth{g})/J$.  The
ideal $J$ is generated by the linear subspace $R \subset
\mathgth{g}^{\otimes2}\oplus\mathgth{g}$ spanned by elements of the form
$hg-gh-[h,g]_{\mathgth{g}}$ for $h\in\mathgth{h},g\in\mathgth{g}$.  Let $qR$ be the image of
$R$ through the projection
$\mathgth{g}^{\otimes2}\oplus\mathgth{g}\to\mathgth{g}^{\otimes2}$, and form the graded
algebra $qA = \mathbf{T}(\mathgth{g})/\langle qR\rangle$, where $\langle qR\rangle$
denotes the two-sided ideal generated by $qR$.  Since $qR$ lies in the
kernel of the quotient algebra morphism $\mathbf{T}(\mathgth{g})\to \gr(A)$, we obtain
a surjective algebra morphism $qA\to \gr(A)$.

The quadratic algebra $qA$ is Koszul (see
Appendix~\ref{sec:appendix}), and for such algebras we can apply a
simple criterion~\cite{BG} to check that the map $qA\rightarrow\gr(A)$ is an
isomorphism.  We describe this result below.

Let $\phi:qR\rightarrow \mathgth{g}$ be the linear map defined as follows. For $x\in
qR$, $\phi(x)$ is the linear part of a preimage of $x$ under the
projection $R\rightarrow qR$.  This is well defined because $R\cap \mathgth{g} =
0$.  Now Theorem~4.1 in~\cite{BG} states that the morphism $qA\to
\gr(A)$ is an isomorphism if and only if the following conditions are
satisfied:
\begin{itemize}
\item[(1)] $\mathsf{Im}~(\phi\otimes{\rm id}-{\rm id}\otimes\phi)\subset qR$
(this map is defined on $qR\otimes\mathgth{g}\cap\mathgth{g}\otimes qR$).
\item[(2)] $\phi\circ(\phi\otimes{\rm id}-{\rm id}\otimes\phi)=0$.
\end{itemize}

In our situation, $qR$ is the vector subspace of $\mathgth{g}\otimes \mathgth{g}$
spanned by $\{hg-gh|h\in\mathgth{h},g\in\mathgth{g}\}$.  The map $\phi$ maps $hg-gh$
to $[h,g]_{\mathgth{g}}$.  Thus condition (2) follows from
the Jacobi identity, while condition (1) is ensured by the stability
of $\mathgth{h}$ under the bracket of $\mathgth{g}$.   We conclude that the map $qA
\rightarrow \gr(A)$ is an isomorphism.

In particular, the $\mathbf{k}$-vector space $A$ can now be identified with
$\mathbf{T}(\mathgth{n})\otimes \S(\mathgth{h})$.  Choose $\mathbf{k}$-linear splittings of the
projections $\mathgth{g}\twoheadrightarrow\mathgth{n}$ and
$\mathbf{T}(\mathgth{h})\twoheadrightarrow\S(\mathgth{h})$.  Then the composed map
\begin{equation}\label{:composed}
  \mathbf{T}(\mathgth{n})\otimes
  \S(\mathgth{h})\hookrightarrow\mathbf{T}(\mathgth{g})\otimes\mathbf{T}(\mathgth{g})\to\mathbf{T}(\mathgth{g})\twoheadrightarrow
  A
\end{equation}
is an isomorphism of filtered $\mathbf{k}$-vector spaces. Applying the
counit of $\S(\mathgth{h})$ then produces a $\mathbf{k}$-linear projection
$\varphi:A\twoheadrightarrow\mathbf{T}(\mathgth{n})$.

Let us now prove that $\ker(\varphi)=A\mathgth{h}$. The kernel of $\varphi$
clearly lies in $A\mathgth{h}$. Conversely, we now prove that any element
$\sum_sa_sh_s\in A\mathgth{h}$ lies in the kernel. For any $s$ we can write
$a_s=\sum_t b_tc_t$ in a unique way with $b_t$, resp.~$c_t$, in the
image of $T(\mathgth{n})$, resp.~$\S(\mathgth{h})$, through \eqref{:composed}. Then
$a_sh_s=\sum_t b_t(c_th_s)\in\ker(\varphi)$ since $c_th_s$ lies in the
augmentation ideal of $\S(\mathgth{h})$.

Therefore we get a filtered isomorphism of $\mathbf{k}$-vector spaces
$\mathbf{T}(\mathgth{n})\tilde\to A/A\mathgth{h}$ obtained as the composed map
$$
\mathbf{T}(\mathgth{n})\hookrightarrow A\twoheadrightarrow A/A\mathgth{h}\,,
$$
where the first inclusion is determined by a $\mathbf{k}$-linear splitting
of $\mathgth{g}\twoheadrightarrow\mathgth{n}$.  We now prove that at the level of
associated graded it respects $\mathgth{h}$-module structures on both sides.
For any $h\in\mathgth{h}$ and any monomial $x_1\cdots x_k\in\mathbf{T}^k(\mathgth{n})$, the
failure of $\mathgth{h}$-linearity is given by
$$
\sum_{i=1}^kx_1\cdots[h,x_i]_{|\mathgth{h}}\cdots x_k
$$
where $[\,,]_{|\mathgth{h}}$ is the $\mathgth{h}$-part of the bracket, which is
defined by means of the above splitting.  We conclude with the very
simple observation that for any $h\in\mathgth{h}$ and any
$x_1,\dots,x_k\in\mathgth{g}$ we have, in $F^{k+1}$,
$$
x_1\cdots h\cdots x_k\in A\mathgth{h}+F^{k}\,.
$$
Therefore the failure of $\mathgth{h}$-linearity vanishes after passing to the
associated graded $\mathgth{h}$-module of $A/A\mathgth{h}$.
\footnote{The very same argument shows that the
  $\mathgth{h}$-module isomorphism $\mathbf{T}^k(\mathgth{n})\to F^k/F^{k-1}$ constructed this
  way does not depend on the choice of a $\mathbf{k}$-linear splitting
  $\mathgth{g}\twoheadrightarrow\mathgth{n}$. }. \qed
\end{Proof}

\paragraph
Next we relate the extension class $\alpha_\mathgth{n}$ with the filtration $F^\cdot$ on $j^*j_!{\mathbf 1}_\mathgth{h}$. The inclusion $F^0\hookrightarrow F^k$ of the ground field always
splits for any $k>0$. We shall denote the reduced filtration by
$\tilde{F}^\cdot$.

By the above lemma we have $\tilde{F}^1 \cong \mathgth{n}$ and
$\tilde{F}^2/\tilde{F}^1 \cong \mathgth{n}^{\otimes 2}$. Hence the inclusion $\tilde{F}^1\hookrightarrow \tilde{F}^2$ defines a short exact sequence
of $\mathgth{h}$-modules
\begin{equation}
\label{ses:4}
0\rightarrow \mathgth{n} \rightarrow \tilde{F}^2 \rightarrow
\mathgth{n}^{\otimes 2} \rightarrow 0 .
\end{equation}
The next lemma shows that the extension class of this sequence is
precisely the class $\alpha := \alpha_\mathgth{n}\in \Ext^1_\mathgth{h}(\mathgth{n}\otimes
\mathgth{n}, \mathgth{n})$ defined in~(\ref{subsec:alpha}).

\begin{Lemma}
\label{lem:3}
The short exact sequences \eqref{ses:3} and \eqref{ses:4} are
isomorphic and hence both define the same obstruction class $\alpha$.
\end{Lemma}

\begin{Proof}
We construct a map between
\[ Q:=\mathgth{n}\oplus (\mathgth{g}\otimes \mathgth{n}) / \langle (h(x),0)-(0,h\otimes
x)\rangle \]
and $\tilde{F}^2$ which makes all the squares commute. The
required map has two components: one from $\mathgth{n}$ and the other from
$\mathgth{g}\otimes \mathgth{n}$.  The first component is the natural
inclusion map $\mathgth{n} = \tilde F^1 \subset \tilde F^2$.
The second one is given by
\[g\otimes x \mapsto [g\otimes \bar{x}] \]
where we first choose a lift $\bar{x}$ of $x\in \mathgth{n}$ to $\mathgth{g}$ and
then take the class of $g\otimes \bar{x} \in \mathbf{T}^2\mathgth{g}$ in $\tilde{F}^2$.  A direct computation checkes that the map is well-defined (independent of lifting) and
that the resulting map factors through $Q$. A quick diagram chasing
shows that all squares commute.  \qed
\end{Proof}

\paragraph
The perhaps surprising result that will be proved in
Proposition~\ref{prop:fon} below is that the vanishing of the
extension class $\alpha$, which by the above lemma is only equivalent
to the splitting of the first nontrivial inclusion $F^1\hookrightarrow
F^2$, is in fact equivalent to the splitting of the entire filtration
$F^\cdot$. We will need the following standard lemma which establishes
an isomorphism of $\mathgth{g}$-modules analogous to the projection formula
in algebraic geometry.

\begin{Lemma}
\label{lem:proj}
Let $i: \mathgth{h}\hookrightarrow \mathgth{g}$ be an inclusion of Lie algebras. Let
$E$ be an $\mathgth{h}$-module and $F$ be a $\mathgth{g}$-module. Then we have an
isomorphism of $\mathgth{g}$-modules
\[ i_!(E)\otimes F \cong i_!(E\otimes i^* F).\]
\end{Lemma}
\vspace{-6mm}

\begin{Proof}
Since the result is well-known to the experts, we only provide a short
outline of its proof.  Let $\Delta:\mathbf{U}(\mathgth{g})\rightarrow \mathbf{U}(\mathgth{g})\otimes
\mathbf{U}(\mathgth{g})$ be the cocommutative coproduct on the universal enveloping
algebra, and let $S:\mathbf{U}(\mathgth{g})\rightarrow \mathbf{U}(\mathgth{g})^{\mathsf{op}}$ be the antipode
map. We shall freely use the sumless Sweedler notation for the coproduct,
\[ \Delta(f)=f^{(1)}\otimes f^{(2)},\  (\Delta\otimes{\rm
  id})\circ\Delta(f)=({\rm
  id}\otimes\Delta)\circ\Delta(f)=f^{(1)}\otimes f^{(2)}\otimes
f^{(3)}, \  \ldots \]
It is then straightforward to check that the linear map
\[ \phi: i_!(E)\otimes
F=(\mathbf{U}(\mathgth{g}) \otimes_{\mathbf{U}(\mathgth{h})} E)\otimes F \rightarrow i_!(E\otimes i^*
F)=\mathbf{U}(\mathgth{g}) \otimes_{\mathbf{U}(\mathgth{h})} (E\otimes F) \]
given by
\[ \phi((f\otimes x) \otimes y)= f^{(1)}\otimes (x\otimes
S(f^{(2)})y)\]
is a well-defined isomorphism, with inverse
\[ \psi: i_!(E\otimes i^*
F)=\mathbf{U}(\mathgth{g}) \otimes_{\mathbf{U}(\mathgth{h})} (E\otimes F) \rightarrow i_!(E)\otimes
F=(\mathbf{U}(\mathgth{g}) \otimes_{\mathbf{U}(\mathgth{h})} E)\otimes F \]
given by
\begin{equation}
\psi(f \otimes(x\otimes y))= (f^{(1)}\otimes x)\otimes
f^{(2)}y. \tag*{\qed}
\end{equation}
\end{Proof}
\vspace{-6mm}

\begin{Proposition}
\label{prop:fon}
The following two statements are equivalent:
\begin{itemize}
\item[(a)] The filtration $F^\cdot$ splits.
\item[(b)] The extension class $\alpha$ is trivial.
\end{itemize}
\end{Proposition}

\begin{Proof}
The implication from (a) to (b) follows from Lemma~\ref{lem:3}. For the other implication we would like to show that the short exact sequences
\[ 0\rightarrow F^{k-1} \rightarrow F^{k} \rightarrow F^{k}/F^{k-1}=\mathgth{n}^{\otimes k}\rightarrow 0\]
split assuming that the extension class $\alpha$ vanishes. Note that the last equality in the above sequences is proved in Lemma~\ref{lem:2}. Below we will explicitly construct $\mathgth{h}$-linear maps $\mathgth{n}^{\otimes k}\rightarrow F^k$ that split the above short exact sequences.

By Lemma~\ref{lem:obs} the condition $\alpha=0$ is equivalent to the existence of a $\mathgth{h}^{(1)}$-module structure on $\mathgth{n}$ that extends the $\mathgth{h}$-module structure on it. Denote by $\overline{\mathgth{n}}$ such an extension. Note that as a vector space $\overline{\mathgth{n}}$ is the same as $\mathgth{n}$. Denote by $\ad$ the structure map $\mathgth{h}^{(1)}\otimes \overline{\mathgth{n}} \rightarrow \overline{\mathgth{n}}$ for the $\mathgth{h}^{(1)}$-module $\overline{\mathgth{n}}$.

We have a natural map of $\mathgth{h}$-modules $\mathgth{n}\hookrightarrow j^*j_! ({\mathbf 1}_\mathgth{h})$ for the inclusion of Lie algebras $j:\mathgth{h}\rightarrow \mathgth{h}^{(1)}$.
By adjunction this defines a map of $\mathgth{h}^{(1)}$-modules
\[ j_!(\mathgth{n}) \rightarrow j_!({\mathbf 1}_\mathgth{h}).\]
 Tensoring both sides with $\overline{\mathgth{n}}$ yields a map $j_!(\mathgth{n})\otimes \overline{\mathgth{n}} \rightarrow j_!({\mathbf 1}_\mathgth{h})\otimes \overline{\mathgth{n}}$.
Applying the projection formula in Lemma~\ref{lem:proj} (for the inclusion $j:\mathgth{h}\hookrightarrow \mathgth{h}^{(1)}$) we get a map
\[ j_! (\mathgth{n}^{\otimes 2}) \rightarrow j_!(n).\]
Iterating this procedure yields for any nonnegative integer $k$ a map of $\mathgth{h}^{(1)}$-modules
\[ j_! (\mathgth{n}^{\otimes k+1}) \rightarrow j_!(\mathgth{n}^{\otimes k}).\]
Hence fixing the integer $k$ we can consider the composition
\[ j_!(\mathgth{n}^{\otimes k})\rightarrow j_!(\mathgth{n}^{\otimes k-1}) \rightarrow \cdots \rightarrow j_!({\mathbf 1}_\mathgth{h}).\]
Applying adjunction to this composition we get a map of $\mathgth{h}$-modules
\[ s_k: \mathgth{n}^{\otimes k} \rightarrow j^*j_!({\mathbf 1}_\mathgth{h}).\]
We need to check that the image of $s_k$ lies inside the $k$-th step
of the filtration $F^\cdot$ and that it splits the natural surjective
map from $F^k$ to $\mathgth{n}^{\otimes k}$ constructed in Lemma~\ref{lem:2}.

By construction the maps $t_{k+1}: j_!(\mathgth{n}^{\otimes k+1}) \rightarrow
j_!(\mathgth{n}^{\otimes k})$ fit into the commutative diagram
\[\begin{CD}
  j_!(\mathgth{n}^{\otimes k+1}) @>t_{k+1}>> j_!(\mathgth{n}^{\otimes k}) \\
  @VV\psi V                                  @AA\phi A \\
  j_!(\mathgth{n}^{\otimes k} )\otimes \overline{\mathgth{n}} @>t_k\otimes \id>>
  j_!(\mathgth{n}^{\otimes k-1})\otimes \overline{\mathgth{n}},\end{CD}\]
where $\psi$ and $\phi$ are the maps defined in the proof of the
projection formula Lemma~\ref{lem:proj}. This inductive construction
begins with the map $t_1:j_! (\mathgth{n}) \rightarrow j_!({\mathbf 1}_\mathgth{h})$ which
is explicitly given by $f\otimes x \mapsto fx \otimes 1$. Hence with
respect to the filtrations induced from $\mathbf{T}(\mathgth{g})$, $t_1$ increases
the filtration degree by $1$.

It is now important to observe that the coproduct used in the
definition of $\phi$ and $\psi$ not only preserves the filtration for
which elements in $\mathgth{h}^{(1)}$ are of degree $1$, but also preserves
the filtration induced from $\mathbf{T}(\mathgth{g})$. This can be seen by observing
that the natural surjective map $\mathbf{T}(\mathgth{g})\rightarrow \mathbf{U}(\mathgth{h}^{(1)})$ is
a morphism of bialgebras (hence in particular a map of
coalgebras). Thus by an induction argument we conclude that the maps
$t_k$ all increase the filtration degree by $1$. The splitting maps
$s_k$ can then be described as the compositions
\[\begin{CD}
  \mathgth{n}^{\otimes k} @>\eta_k>> j^*j_!(\mathgth{n}^{\otimes k}) @>j^*t_k>>
  j^*j_!(\mathgth{n}^{\otimes k-1}) @>>> \cdots @>>>
  j^*j_!({\mathbf 1}_\mathgth{h}).\end{CD}\]
Here $\eta_k$ is the unit of the adjunction applied to $\mathgth{n}^{\otimes
  k}$, explicitly given by $ x_1\otimes\cdots\otimes x_k\mapsto
1\otimes_{\mathbf{U}(\mathgth{h})} (x_1\otimes\cdots\otimes x_k)$. As $\eta_k$
decreases the filtration by $k$ and we have $k$ times $t$'s to
post-compose with it, the map $s_k$ will end up being a filtration
preserving map, i.e., its image lies inside $F^k\subset
j^*j_!({\mathbf 1}_\mathgth{h})$.

A direct computation of the map $s_k$ shows that it splits the surjective map $F^k\rightarrow \mathgth{n}^{\otimes k}$. For instance in the cases $k=2$ and $k=3$, we have
\begin{itemize}
\item[k=2]   $x_1\otimes x_2 \mapsto x_1 x_2 - \ad(x_1)x_2.$
\item[k=3]   $x_1\otimes x_2\otimes x_3 \mapsto x_1x_2x_3-x_1\ad(x_2)x_3-x_2\ad(x_1)x_3+\ad(x_2)\ad(x_1)x_3- \ad(x_1)x_2x_3+\ad(\ad(x_1)x_2) x_3.$
\end{itemize}
Lifting is assumed in this formula whenever necessary. One can check
directly that that this formula is independent of all liftings
involved. However it does depend on the choice of the
$\mathgth{h}^{(1)}$-module $\bar{\mathgth{n}}$ which lifts the $\mathgth{h}$-module structure
on $\mathgth{n}$. \qed
\end{Proof}

\section{First order neighborhood Lie algebras}
\label{sec:hkr}

Let $\mathgth{h}\hookrightarrow \mathgth{g}$ be an inclusion of Lie algebras and
denote by $\mathgth{n}$ the quotient $\mathgth{h}$-module $\mathgth{g}/\mathgth{h}$.  In this section
we define the obstruction class $\alpha$ and the first order neighborhood Lie algebra $\mathgth{h}^{(1)}$ that appear in Theorem~\ref{thm:mainthm}.  Then we prove that the class $\alpha$ is trivial if and only if $\mathgth{n}$ admits an extension to $\mathgth{h}^{(1)}$.

\paragraph
\label{subsec:alpha}
\textbf{The extension class $\alpha$.} We begin with the definition of
the extension class $\alpha$ that appears in the statement of
Theorem~\ref{thm:mainthm}.  Consider the short exact sequence of
$\mathgth{h}$-modules
\begin{equation}
\label{ses:1} 0\rightarrow \mathgth{h} \rightarrow \mathgth{g} \rightarrow \mathgth{n}
\rightarrow 0.
\end{equation}
Let $E$ be an $\mathgth{h}$-module.  Tensoring \eqref{ses:1} with $E$ yields
the sequence
\begin{equation}
\label{ses:2} 0\rightarrow \mathgth{h}\otimes E \rightarrow \mathgth{g} \otimes E
\rightarrow \mathgth{n}\otimes E \rightarrow 0
\end{equation}
which remains exact because the tensor product of representations is
the tensor product of vector spaces endowed with the $\mathgth{h}$-module
structure given by the Leibniz rule.  The extension class
of~(\ref{ses:2}) is a map $\mathgth{n}\otimes E \rightarrow \mathgth{h}\otimes E[1]$
in the derived category of $\mathgth{h}$-representations, which can be
post-composed with the action map $\mathgth{h}\otimes E\rightarrow E$ to give
the map
\[ \alpha_E:\mathgth{n}\otimes E \rightarrow E[1]. \]
Equivalently, we can define $\alpha_E$ as the class in
$\Ext^1_\mathgth{h}(\mathgth{n}\otimes E, E)$ corresponding to the bottom extension in
the diagram below:
\begin{equation}
\label{ses:3}
\begin{CD} 0 @>>> \mathgth{h}\otimes E @>>> \mathgth{g}\otimes E @>>> \mathgth{n}\otimes E
@>>> 0 \\ @.  @VVV @VVV @| @. \\ 0 @>>> E @>>> Q @>>> \mathgth{n}\otimes E
@>>> 0.
\end{CD}
\end{equation}
Here the $\mathgth{h}$-module $Q$ is obtained by push-out in the first square
of the above diagram.  Explicitly, it is given by
\[ Q= E\oplus (\mathgth{g}\otimes E) / \langle (h(x),0)-(0,h\otimes
x)\rangle \]
where for $h\in\mathgth{h}$ and $x\in E$ we have denoted by $h(x)$ the action
of $h$ on $x$ and $h\otimes x$ is viewed as an element of $\mathgth{g}\otimes
E$ via the inclusion of $\mathgth{h}$ into $\mathgth{g}$.

We will be particularly interested in the class $\alpha_\mathgth{n}$ of
the $\mathgth{h}$-module $\mathgth{n}$.  This special class will be denoted simply by $\alpha$.

\paragraph
\textbf{The first order neighborhood Lie algebra $\mathgth{h}^{(1)}$.}
Consider the Lie algebra $\mathgth{h}^{(1)}$ defined by
\[ \mathgth{h}^{(1)} := \mathbf{L}(\mathgth{g})/\langle [h,g]-[h,g]_\mathgth{g}~|~h\in\mathgth{h}, g\in
\mathgth{g} \rangle \]
where $\mathbf{L}(\mathgth{g})$ denotes the free Lie algebra generated by the vector
space $\mathgth{g}$ and $\langle\rangle$ stands for ``Lie ideal generated by''.
More precisely $\mathgth{h}^{(1)}$ is the quotient of $\mathbf{L}(\mathgth{g})$ in
which the bracket between elements of $\mathgth{h}$ and $\mathgth{g}$ has been
identified with the original one in $\mathgth{g}$. Note that to define the
Lie algebra $\mathgth{h}^{(1)}$ we do not need $\mathgth{g}$ to be a Lie algebra. The
precise weaker condition for which this construction makes sense is given in
Lemma~\ref{lem:obs} below.

There are natural maps of Lie algebras
\[ \mathgth{h} \hookrightarrow \mathgth{h}^{(1)} \mbox{ and } \mathgth{h}^{(1)}\rightarrow
\mathgth{g}\]
which factor the original inclusion $\mathgth{h}\hookrightarrow \mathgth{g}$.  Given
an $\mathgth{h}$-representation $E$ we can ask whether $E$ extends to a
representation of $\mathgth{h}^{(1)}$. In other words we ask if on the vector
space $E$ we can find an $\mathgth{h}^{(1)}$-module structure whose
restriction to $\mathgth{h}$ via the map $\mathgth{h}\rightarrow \mathgth{h}^{(1)}$ is the original
one.  The following lemma shows that this is the case if and only if
$\alpha_E=0$. We state the lemma in a slightly greater generality.

\begin{Lemma}
\label{lem:obs}
Let $\mathgth{h}$ be a Lie algebra and let $\mathgth{g}$ be an $\mathgth{h}$-module that
contains $\mathgth{h}$ as an $\mathgth{h}$-submodule. An $\mathgth{h}$-module $E$ is the
restriction of an $\mathgth{h}^{(1)}$-module if and only if its class
$\alpha_E$ is trivial.
\end{Lemma}

\begin{Proof}
We begin with the if part. Assume that the class $\alpha_E$ is
trivial. This implies that the sequence \eqref{ses:3} splits in the
category of $\mathgth{h}$-modules. Thus we get a map $j:Q\rightarrow E$ of
$\mathgth{h}$-modules that splits the canonical map $E\rightarrow Q$. Pre-composing
$j$ with the middle vertical map in \eqref{ses:3} yields a map of
$\mathgth{h}$-modules
\[ \rho: \mathgth{g}\otimes E \rightarrow E.\]
This map does not define a representation of $\mathgth{g}$ on $E$, but it
certainly defines a representation of $\mathbf{L}(\mathgth{g})$ by the universal
property of $\mathbf{L}(\mathgth{g})$. The fact that $\rho$ respects the $\mathgth{h}$
structure translates into the fact that $\langle
[h,g]-[h,g]_\mathgth{g}\rangle $ is in the kernel of this
representation. Thus $\rho$ gives an $\mathgth{h}^{(1)}$-module structure on
$E$ which lifts the original $\mathgth{h}$-module structure because the first
square in \eqref{ses:3} commutes.

For the only if part assume we have an $\mathgth{h}^{(1)}$-module structure on
$E$ that lifts the $\mathgth{h}$ structure. Again denote this
action by $\rho$. We can use the explicit description of $Q$
above to define a splitting
\[ (x,g\otimes y)\mapsto (x+\rho(g)(y)) .\]
This map is obviously a splitting and it respects the $\mathgth{h}$-module
structure because $\langle [h,g]-[h,g]_\mathgth{g}\rangle$ is in the kernel
of the representation $\rho$.  \qed
\end{Proof}

\section{Introduction}
\label{sec:intro}

\subsection{The aim}

Let $\mathgth{h}\hookrightarrow \mathgth{g}$ be an inclusion of Lie algebras.  Denote
by $\mathgth{n}$ the quotient $\mathgth{g}/\mathgth{h}$.  The quotient $\mathbf{U}(\mathgth{g})/\mathbf{U}(\mathgth{g})
\mathgth{h}$ of $\mathbf{U}(\mathgth{g})$ by the left ideal generated by $\mathgth{h}$ is naturally
an $\mathgth{h}$-representation.  The main purpose of this paper is to answer
the following question (the PBW problem):

\begin{center}
  \textit{When is $\mathbf{U}(\mathgth{g})/\mathbf{U}(\mathgth{g}) \mathgth{h}$ isomorphic to $\mathbf{S}(\mathgth{n})$ as
    $\mathgth{h}$-representations?}
\end{center}

\noindent
A more precise way of stating the above question is the following. The representation ${\mathbf U}(\gog)/{\mathbf U}(\gog)\mathgth{h}$ admits a natural filtration by
$\mathgth{h}$-modules whose associated graded $\mathgth{h}$-module is $\S(\mathgth{n})$.  We
ask for a necessary and sufficient condition for this filtration to
split.

This question is important in deformation quantization, as the space
of $\mathgth{h}$-invariants $\left ( {\mathbf U}(\gog)/{\mathbf U}(\gog)\mathgth{h} \right )^\mathgth{h}$ can be given a
natural structure of algebra by identifying it with the space of
invariant differential operators on a homogeneous space~\cite{Tor}.
An open conjecture of Duflo is concerned with understanding the center
of this algebra in terms of the Poisson center of $\S(\mathgth{n})^\mathgth{h}$, which
is thought of as the algebra of functions on a Poisson manifold
obtained via reduction through the moment map $\mathgth{g}^{\scriptscriptstyle\vee} \rightarrow
\mathgth{h}^{\scriptscriptstyle\vee}$.  In order for this conjecture to make sense one needs to be
in a situation where the PBW isomorphism holds.  Traditionally this is
achieved by assuming that the inclusion $\mathgth{h} \hookrightarrow \mathgth{g}$
splits as a map of $\mathgth{h}$-modules.  We will see that this condition is
unnecessarily restrictive: there are many pairs of Lie algebras for
which there is a PBW isomorphism (and hence it makes sense to study
the Duflo problem), but for which the inclusion $\mathgth{h}\hookrightarrow
\mathgth{g}$ does not split.

\subsection{An analogous problem in algebraic geometry}

Kontsevich and Kapranov~\cite{Ka} had the insight that we can view the
shifted tangent sheaf $T_Y[-1]$ of a smooth algebraic variety
$Y$ as a Lie algebra object in the derived category ${\mathbf D}(Y)$ of
coherent sheaves on $Y$, with bracket given by the Atiyah class of the
tangent sheaf.  Moreover, the Atiyah class of any object in ${\mathbf D}(Y)$
gives it the structure of module over this Lie algebra object (see for
example~\cite{RW}).  Loosely speaking ${\mathbf D}(Y)$ can be regarded as the
category of representations of the shifted tangent sheaf.  The role of
the trivial representation is played by the structure sheaf ${\mathscr O}_Y$.

An embedding $i:X\hookrightarrow Y$ of smooth algebraic varieties can
be thought of as giving rise to an inclusion of Lie algebra objects in
${\mathbf D}(X)$
\[ \mathfrak h=T_X[-1]\hookrightarrow i^*T_Y[-1]=\mathfrak{g}. \]
If $E$ is an object in ${\mathbf D}(Y)$ then the Atiyah class of the
restriction $i^*E$ of $E$ to $X$ is precisely the composite of the
above inclusion of Lie algebras with the restriction to $X$ of the
Atiyah class of $E$.  In other words the functor
\[ i^*:{\mathbf D}(Y) \rightarrow {\mathbf D}(X) \]
can be interpreted as the restriction functor
\[ \Res:\mathgth{g}\mbox{-}\mathgth{Mod} \rightarrow \mathgth{h}\mbox{-}\mathgth{Mod}. \]
(We think of all our functors between derived categories as being
implicitly derived, so we write $i^*$ instead of $\mathbf{L} i^*$, etc.)

We now see a dictionary emerging between the worlds of Lie theory and
of algebraic geometry. We can use this dictionary to translate naively
the PBW question into a problem in algebraic geometry.  The following
concepts are matched by this dictionary:

\begin{center}
\begin{tabular}{l|l}
Lie theory & Algebraic geometry\\
\hline
Lie algebras $\mathgth{h}$, $\mathgth{g}$ & varieties $X$, $Y$, $\mathgth{h} = T_X[-1]$,
$\mathgth{g} = T_Y[-1]$ \\
inclusion $\mathgth{h}\hookrightarrow \mathgth{g}$ & closed embedding
$i:X\hookrightarrow Y$ \\
$\mathgth{h}\mbox{-}\mathgth{Mod}$, $\mathgth{g}\mbox{-}\mathgth{Mod}$ & ${\mathbf D}(X)$, ${\mathbf D}(Y)$ \\
${\mathbf 1}_\mathgth{h}\in \mathgth{h}\mbox{-}\mathgth{Mod}$ & ${\mathscr O}_X \in {\mathbf D}(X)$ \\
$\Res:\mathgth{g}\mbox{-}\mathgth{Mod} \rightarrow \mathgth{h}\mbox{-}\mathgth{Mod}$ & $i^*: {\mathbf D}(Y) \rightarrow {\mathbf D}(X)$
\\
$\Ind:\mathgth{h}\mbox{-}\mathgth{Mod} \rightarrow \mathgth{g}\mbox{-}\mathgth{Mod}$ & $i_!:{\mathbf D}(X) \rightarrow {\mathbf D}(Y)$
\end{tabular}
\end{center}

\noindent
The last line is motivated by the fact that the induction functor
$\Ind$ is the left adjoint of the restriction functor, hence in the
right column we take the left adjoint $i_!$ of the pull-back functor,
which exists for a closed embedding $i$ of smooth varieties.

In representation-theoretic language the $\mathgth{h}$-representation
${\mathbf U}(\gog)/{\mathbf U}(\gog)\mathgth{h}$ arises as
\[ {\mathbf U}(\gog)/{\mathbf U}(\gog)\mathgth{h} = \Res \Ind {\mathbf 1}_\mathgth{h} \in \mathgth{h}\mbox{-}\mathgth{Mod}. \]
Using the dictionary the latter corresponds to the object $i^* i_!
{\mathscr O}_X$ of the derived category ${\mathbf D}(X)$.  Any object $E$ of ${\mathbf D}(X)$
admits a natural filtration by successive truncations $\tau^{\geq k}
E$ whose $k$-th ``quotient'' is the cohomology sheaf ${\mathscr H}^k(E)[-k]$.
An easy local calculation shows that for $E = i^* i_! {\mathscr O}_X$ we have
\[ {\mathscr H}^k(i^* i_! {\mathscr O}_X) = \wedge^k N \]
where $N$ is the normal bundle of $X$ in $Y$.  Thus the associated
graded object of $i^*i_! {\mathscr O}_X$ is precisely
\[ \gr (i^* i_! {\mathscr O}_X) = \bigoplus_k \wedge^k N[-k] = \S(N[-1]). \]
Since $N[-1] = T_Y[-1]|_X/T_X[-1]$ corresponds via the dictionary to
$n= \mathgth{g}/\mathgth{h}$, this is the precise analogue of the statement that
${\mathbf U}(\gog)/{\mathbf U}(\gog)\mathgth{h}$ admits a filtration whose associated graded is
\[ \gr \left ({\mathbf U}(\gog)/{\mathbf U}(\gog)\mathgth{h} \right ) = \S(n). \]

The PBW question translates into the following question about a
closed embedding $i:X\hookrightarrow Y$.
\begin{center}
\textit{When is $i^*i_!{\mathscr O}_X$ isomorphic to $\S(N[-1])$ in
  ${\mathbf D}(X)$?}
\end{center}
Just like in the usual PBW problem, this question is better phrased by
asking when the above filtration on $i^*i_!{\mathscr O}_X$ splits.  This
question was addressed and solved recently by D.~Arinkin and the second
author in~\cite{AC}, where they prove the following result.
\medskip

\label{thm:alggeom}
\noindent \textbf{Theorem~1.2.}
{\em
Let $X^{(1)}$ be the first infinitesimal neighborhood of $X$ in $Y$,
$X\hookrightarrow X^{(1)}$.  The following are equivalent:
\medskip
\begin{enumerate}
\item the truncation filtration on $i^*i_!\mathcal O_X$ splits, giving
  rise to an isomorphism
  \[ i^*i_! {\mathscr O}_X \cong \S(N[-1]); \]
\item the class $\alpha$ is trivial, where
  \[ \alpha\in\Ext^1_X(N[-1]^{\otimes 2},N[-1]) \]
  is obtained by composing the class of the normal bundle exact sequence
  with the Atiyah class of the normal bundle $N$;
\item the vector bundle $N[-1]$ admits an extension to $X^{(1)}$.
\end{enumerate}
}
\medskip

\noindent
It is worth noting that there are many cases where the short exact sequence
\[ 0 \rightarrow T_X \rightarrow T_Y|_X \rightarrow N \rightarrow 0\]
does not split but the obstruction $\alpha$ is nonetheless trivial.
For example this is the case when $X$ is any non-linear hypersurface
in $Y = \mathbf{P}^n$.

\subsection{The result}

Our main result is the following translation of the above theorem.
\medskip

\noindent\label{thm:mainthm}{\bf Theorem~1.3.}
{\em
  There exists a Lie algebra $\mathgth{h}^{(1)}$, containing $\mathgth{h}$ as a
  Lie subalgebra, such that the following are equivalent:
\medskip
\begin{enumerate}
\item the natural filtration on ${\mathbf U}(\gog)/{\mathbf U}(\gog) \mathgth{h}$ splits, giving rise to an isomorphism of $\mathgth{h}$-modules
  \[ {\mathbf U}(\gog)/{\mathbf U}(\gog)\mathgth{h}\cong \S(\mathgth{n}); \]
\item the class $\alpha$ is trivial, where
  $\alpha\in\Ext^1_\mathgth{h}(\mathgth{n}^{\otimes2},\mathgth{n})$ is obtained by composing the class of
\[ 0\longrightarrow \mathfrak h\longrightarrow\mathfrak
g\longrightarrow\mathfrak n\longrightarrow 0 \]
with the $\mathgth{h}$-action;
\item the $\mathgth{h}$-representation $\mathgth{n}$ admits an extension to $\mathgth{h}^{(1)}$.
\end{enumerate}
}
\medskip

\noindent
Observe that in the algebro-geometric context $X^{(1)}$ is singular,
even though $X$ and $Y$ are smooth. It turns out that the correct
notion of tangent bundle for such singular spaces is that of {\em
  tangent complex}, see~\cite{Illusie}.  Consequently the Lie algebra
$\mathgth{h}^{(1)}$ shall be defined following the analogy with the tangent
complex of $X^{(1)}$, using insight from Koszul duality.  The details
are presented in Section~\ref{sec:hkr}.

The paper is organized as follows. Section~\ref{sec:hkr} is devoted to
the definition of the ``first order neighborhood Lie algebra''
$\mathgth{h}^{(1)}$ and to the proof that an $\mathgth{h}$-module $E$ admits an
extension to $\mathgth{h}^{(1)}$ if and only if a certain class $\alpha_E\in
Ext^1(\mathgth{n}\otimes E,E)$ is trivial.  In Section 3 we prove a variant of
our main theorem for the inclusion $\mathgth{h}\hookrightarrow\mathgth{h}^{(1)}$.
More precisely, we prove that a natural filtration on
$U(\mathgth{h}^{(1)})/U(\mathgth{h}^{(1)})\mathgth{h}$ splits if and only if the class
$\alpha:=\alpha_\mathgth{n}$ is trivial.  The following section is concerned
with proving Theorem~\ref{thm:mainthm}.  While the main theorem is
concerned with the study of the induction-restriction of the trivial
representation, we can deduce from this case a general result for any
$\mathgth{h}$-representation.  A sketch of the general case is contained in
Section~\ref{sec:twisted}.  We conclude the paper with a very short
section in which we give a simple example of a pair of Lie algebras
for which the class $\alpha$ is non-trivial.
Appendix~\ref{sec:appendix} contains a proof of the fact that a
certain algebra used in the definition of $\mathgth{h}^{(1)}$ is Koszul.
\medskip

\noindent
\textbf{Assumptions.}
In what follows all algebraic structures are considered over a given
field $\mathbf{k}$.  For the main result we need to assume that the
characteristic of $\mathbf{k}$ is zero, but all other results hold without
this assumption.
\medskip

\noindent
\textbf{Acknowledgements.} We are grateful to D.\ Grinberg for his
careful reading of the paper and for pointing out two important
mistakes in a previous version.  Discussions with D.\ Grinberg, G.\
Felder, and C.\ Rossi helped improve the exposition and clarify the
statement and proof of Theorem~\ref{thm:twisted}. We also extend
thanks to M.\ Duflo and C.\ Rossi who commented on early versions of
the result and provided the encouragement to write this paper.  The
second and third authors were partially supported by the National
Science Foundation under Grant No. DMS-0901224.

\section{PBW for inclusions of Lie algebras}
\label{sec:pbw}

In this section we prove the main result of this paper,
Theorem~\ref{thm:mainthm}.  Explicitly, we show that the filtration on
$\mathbf{U}(\mathgth{g})/\mathbf{U}(\mathgth{g})\mathgth{h}$ splits if and only if the class $\alpha$
vanishes.

\paragraph
We shall concentrate our attention on the $\mathgth{h}$-representation
\[ i^*i_!({\mathbf 1}_\mathgth{h})=\mathbf{U}(\mathgth{g})/\mathbf{U}(\mathgth{g})\mathgth{h}. \]
This module can be realized as the quotient
$\mathbf{T}(\mathgth{g})/\left(I+\mathbf{T}(\mathgth{g})\mathgth{h}\right)$ where $I$ is the two-sided ideal
generated by
\[ \{ g_1\otimes g_2-g_2\otimes g_1-[g_1,g_2]~|~g_1, g_2\in \mathgth{g} \} \,.\]
We denote by
\[ R^0\subset R^1\subset\cdots\subset R^k\subset \cdots \]
the filtration by $\mathgth{h}$-submodules of $i^*i_!({\mathbf 1}_\mathgth{h})$ induced from
the degree filtration on $\mathbf{T}(\mathgth{g})$. We set $G^k:=R^k/R^{k-1}$.

\paragraph
Consider the map
\[ j^*j_!({\mathbf 1}_\mathgth{h})\rightarrow j^*k^*k_!j_!({\mathbf 1}_\mathgth{h})=i^*i_!({\mathbf 1}_\mathgth{h}) \]
constructed using the unit map of the adjunction $k_!\dashv
k^*$. This map preserves the filtrations and descends to maps between
associated graded $\mathgth{h}$-modules
\[ \tau: \mathbf{T}(\mathgth{n}) = \gr(j^*j_! ({\mathbf 1}_\mathgth{h})) \rightarrow
\gr(i^*i_!({\mathbf 1}_\mathgth{h})). \]
From the descriptions of $j^*j_!({\mathbf 1}_\mathgth{h})$ and $i^*i_!({\mathbf 1}_\mathgth{h})$ via quotients of $\mathbf{T}(\mathgth{g})$ we see that the map $\tau$ is surjective.

\begin{Lemma}
\label{lem:4} The kernel of the map $\tau$ is generated by the
commutators $x\otimes y -y\otimes x$ for $x,y\in\mathgth{n}$.
\end{Lemma}

\begin{Proof} The idea is to use the classical ${\mathsf{PBW}}$ theorem for a single Lie algebra.
Consider the increasing filtration $E^0\subset\cdots\subset
E^k\subset\cdots$ on the universal enveloping algebra $\mathbf{U}(\mathgth{g})$. The
classical ${\mathsf{PBW}}$ theorem asserts that the kernel of the canonical
surjective map $\mathgth{g}^{\otimes k}\rightarrow E^k/E^{k-1}$ is generated by the
commutators of elements in $\mathgth{g}$, thus yielding an isomorphism
between the $k$-th symmetric tensors on $\mathgth{g}$ and $E^k/E^{k-1}$.

As all these filtrations are compatible (they all arise from the degree
filtration on $\mathbf{T}(\mathgth{g})$), the surjective map $j^*j_!({\mathbf 1}_\mathgth{h})\rightarrow i^*i_!({\mathbf 1}_\mathgth{h})$ induces surjections
on the associated graded to give maps $\mathgth{n}^{\otimes k} \rightarrow G^k$. Consider the following commutative diagram
\[\begin{CD} 0 @>>> I_1 @>>> \mathgth{g}^{\otimes k} @>>> E^k/E^{k-1} @>>> 0
\\ @.  @VVV @VVV @VVV @. \\ 0 @>>> I_2 @>>> \mathgth{n}^{\otimes k} @>\tau^k>> G^k
@>>> 0
\end{CD}\]
where $I_1$ is the degree $k$ part of the commutator ideal in
$\mathbf{T}(\mathgth{g})$ by the ${\mathsf{PBW}}$ theorem and $I_2$ is the kernel of the
map $\mathgth{n}^{\otimes k} \rightarrow G^k$.

We want to show that $I_2$ is the $k$-th commutator in $\mathgth{n}$. It
suffices to show that the map $I_1\rightarrow I_2$ is surjective. By
the snake lemma this is equivalent to showing that the map from the
kernel of $\mathgth{g}^{\otimes k}\rightarrow \mathgth{n}^{\otimes k}$ to the kernel
of $E^k/E^{k-1}\rightarrow G^k$ is surjective.

For that we consider another commutative diagram:
\[\begin{CD} 0 @>>> E^{k-1} @>>> E^k @>>> E^k/E^{k-1} @>>> 0 \\ @.
@VVV @VVV @VVV @. \\ 0 @>>> R^{k-1} @>>> R^k @>>> G^k @>>> 0.
\end{CD}\]
Since all the vertical maps are surjective, the snake lemma shows that we have a surjection
from the kernel of $E^k\rightarrow R^k$ to the kernel of $E^k/E^{k-1}\rightarrow G^k$. But the kernel of $E^k\rightarrow R^k$ is the right ideal
generated by $\mathgth{h}$ in $\mathbf{U}(\mathgth{g})$ intersected with $E^k$, which is a subset of the kernel of $\mathgth{g}^{\otimes k}\rightarrow \mathgth{n}^{\otimes k}$. Thus the kernel of $\mathgth{g}^{\otimes k}\rightarrow \mathgth{n}^{\otimes k}$ also surjects onto the kernel of $E^k/E^{k-1}\rightarrow G^k$. Thus the lemma is proved.  \qed
\end{Proof}
\medskip

\noindent
To state an if and only if condition for the ${\mathsf{PBW}}$ property of inclusions of Lie algebras, we need the following lemma concerning the obstruction class $\alpha$.

\begin{Lemma}
\label{lem:5} The obstruction class $\alpha\in \Ext^1(\mathgth{n}\otimes \mathgth{n},
\mathgth{n})$ factors through $\S^2(\mathgth{n})$.
\end{Lemma}

\begin{Proof}
The lemma can be seen as a corollary of Lemma~\ref{lem:3} and
Lemma~\ref{lem:4}. Indeed, by Lemma~\ref{lem:3}, we can consider the
following commutative diagram:
\[\begin{CD} 0@>>> \mathgth{n} @>>> \tilde{F}^2 @>>> \mathgth{n}\otimes \mathgth{n} @>>> 0\\
@.  @VVV @VVV @VVV @. \\ 0@>>> \mathgth{n} @>>> \tilde{R}^2 @>>> G^2 @>>> 0
\end{CD}\]
where the vertical maps are all defined via the adjunction
$j^*j_!({\mathbf 1}_\mathgth{h})\rightarrow i^*i_!({\mathbf 1}_\mathgth{h})$. By Lemma~\ref{lem:4}
$G^2 = \S^2(\mathgth{n})$ and the last vertical map is the canonical quotient
from the tensor product to the symmetric product. Direct calculation shows that the second square is Cartesian. Thus
the lemma is proved.  \qed
\end{Proof}
\medskip

\noindent
We can summarize our main result in the following theorem.

\begin{Theorem}
Let $\mathbf{k}$ be a field and let $\mathgth{h}\hookrightarrow \mathgth{g}$ be an inclusion
of Lie algebras over $\mathbf{k}$. Consider the two filtrations $R^0\subset
R^1\subset\cdots\subset R^k\subset \cdots $ and $F^0\subset
F^1\subset\cdots\subset F^k\subset \cdots $ defined above. We have:
\begin{itemize}
\item[--] $\gr(F^\cdot) = \mathbf{T}(\mathgth{n})$;
\item[--] $\gr(R^\cdot) = \S(\mathgth{n})$.
\end{itemize}
Moreover, if the field $\mathbf{k}$ has characteristic zero, then
the following are equivalent:
\begin{itemize}
\item[(a)] The extension class $\alpha$ is trivial.
\item[(b)] The filtration $F^0\subset F^1\subset\cdots\subset
F^k\subset \cdots $ splits;
\item[(c)] The filtration $R^0\subset R^1\subset\cdots\subset
R^k\subset \cdots $ splits.
\end{itemize}
In fact, if the extension class $\alpha$ is trivial, we have the
following explicit splitting of the filtration $R^\cdot$ that resembles the
standard ${\mathsf{PBW}}$ isomorphism:
\[ I: \S(\mathgth{n})\rightarrow \mathbf{T}(\mathgth{n})\cong j^*j_!({\mathbf 1}_\mathgth{h})
\rightarrow i^*i_!({\mathbf 1}_\mathgth{h})\cong \mathbf{U}(\mathgth{g})/\mathbf{U}(\mathgth{g})\mathgth{h}. \]
Here the first arrow is given by any (graded) splitting of the surjective morphism $\mathbf{T}(\mathgth{n})\to\S(\mathgth{n})$ in $\mathgth{h}$-$\mathfrak{Mod}$.
\end{Theorem}

\begin{Proof}
The fact that $\gr(F^\cdot)=\mathbf{T}(\mathgth{n})$ is proved in Lemma~\ref{lem:2}, and $\gr(R^\cdot)=\S(\mathgth{n})$ follows from Lemma~\ref{lem:4}. For the second part of the theorem,
Proposition~\ref{prop:fon} shows that (a) and (b) are equivalent. By Lemma~\ref{lem:5} (c) implies (a). Below we will show that (b) implies (c) and hence
all of (a), (b), (c) are equivalent.

Assuming a splitting $I$ of the surjection $\mathbf{T}(\mathgth{n})\rightarrow
\S(\mathgth{n})$ (which always exists over a field of characteristic zero) and a splitting $s$ of the filtration $F^\cdot$,
we can define the following composition
\[ \S^k(\mathgth{n}) \stackrel{I^k}{\rightarrow} \mathbf{T}^k(\mathgth{n}) \stackrel{s^k}{\rightarrow} F^k \rightarrow R^k.\]
Here the last map is the canonical surjective map from $F^k$ to $R^k$. The fact that this composition defines a splitting for the filtration $R^\cdot$ follows
from the following commutative diagram
\[\begin{CD} \S^k(\mathgth{n}) @>I^k>> \mathbf{T}^k(\mathgth{n}) = F^k/F^{k-1}
@>s^k>> F^k\\ @.  @VVV
@VVV \\ @.  \S^k(\mathgth{n}) = R^k/R^{k-1} @<<< R^k.
\end{CD}\]
\vspace{-2.37em}

\qed
\end{Proof}

\section{Generalization to any representation}
\label{sec:twisted}

The main goal of this section is to extend Theorem~\ref{thm:mainthm}
from the case of the trivial representation ${\bf 1}_\mathgth{h}$ to that of
an arbitrary finite dimensional $\mathgth{h}$-representation $V$.  Consider
the filtrations $R^0\subset R^1\subset\cdots R^k\subset\cdots$,
$F^0\subset F^1\subset\cdots F^k\subset\cdots$, on $i^*i_!(V)$,
$j^*j_!(V)$, respectively, which are induced by the degree filtration
on $T(\mathgth{g})\otimes V$.  Then we have the following theorem.

\begin{Theorem}
\label{thm:twisted}
There are isomorphisms of $\mathgth{h}$-modules $gr(F^*)=T(\mathgth{n})\otimes V$ and
$gr(R^*)=S(\mathgth{n})\otimes V$.  Moreover, the following are equivalent:
\begin{itemize}
\item[(a)] The extension classes $\alpha$ and $\alpha_V$ are trivial.
\item[(b)] The filtration $F^0\subset F^1\subset\cdots
  F^k\subset\cdots$ splits.
\item[(c)] The filtration $R^0\subset R^1\subset\cdots
  R^k\subset\cdots$ splits.
\end{itemize}
\end{Theorem}

\noindent\emph{Sketch of proof.}
First observe that as vector spaces we have filtered $\mathbf{k}$-linear
isomorphisms
$$
i^*i_!(V)\cong i^*i_!({\bf 1}_\mathgth{h})\otimes V\quad\textrm{and}\quad j^*j_!(V)\cong j^*j_!({\bf 1}_\mathgth{h})\otimes V
$$
They are not isomorphisms of $\mathgth{h}$-modules, but on the level of associated graded they induce $\mathgth{h}$-module isomorphisms.
This proves the first part of the theorem.

Contrary to the situation when the representation is trivial, the
inclusions $F^0\hookrightarrow F^k$ and $R^0\hookrightarrow R^k$ do
not automatically split. In particular the inclusion
$V=F^0=R^0\hookrightarrow F^1=R^1$ splits if and only if $\alpha_V$ is
trivial.

Finally, if $\alpha_V$ is trivial then there exists an $\mathgth{h}^{(1)}$-module $\tilde{V}$ such that
$Res(\tilde{V})=V$. From this we deduce an isomorphism of $\mathgth{h}$-modules
$j^*j_!(V)\cong j^*j_!({\bf 1}_\mathgth{h})\otimes V$. We conclude by using the fact that the Theorem is true
for the trivial representation. \qed

redpajama_set_name: RedPajamaArXiv


text:
require File.expand_path(File.dirname(__FILE__) + "/../spec_helper")

describe :externals do
  before(:all) do
    @stdout = $stdout
    $stdout = StringIO.new
    # Test $stdout.string
  end

  after(:all) do
    $stdout = @stdout
  end

  describe "help" do
    it "should show a help message with no input" do
      run
      $stdout.string.should match(/Usage: externals/)
    end
  end

  describe "status" do
    it "should tell the user if the repository exists" do
    end

    it "should tell the user if the repository is frozen" do
    end

    it "should tell the user if a snapshot exists" do
    end
  end

  describe "freeze" do
    describe "on a non-existent repository" do
      it "should clone a new repository" do
      end
    end

    describe "on a frozen repository" do
      it "should tell the user the repository is already frozen" do
      end
    end

    describe "on an unfrozen repository" do
      it "should make the temp directory if it does not exist" do
      end

      it "should compress the repository's .git folder to the temp directory" do
      end

      it "should remove the repository's .git folder" do
      end

      describe "when a snapshot exists" do
        it "should ask the user if the user wants to overwrite the snapshot" do
        end

        it "should not overwrite if the user says no" do
        end
      end
    end
  end

  describe "unfreeze" do
    describe "on a frozen repository" do
      it "should tell the user the repository is already unfrozen" do
      end
    end

    describe "on a non-existent repository" do
      it "should clone the repository" do
      end
    end

    describe "on a frozen repository" do
      describe "that has a snapshot" do
        it "should decompress the snapshot" do
        end

        it "should move the snapshot back to the repository" do
        end

        it "should remove the snapshot" do
        end

        it "should reset the repository" do
        end
      end

      describe "that does not have a snapshot" do
        it "should clone a new repository" do
        end
      end
    end
  end

  def run(action=nil)
    app = Externals::App.new(FileUtils.pwd)
    app.run(action)
  end
end

redpajama_set_name: RedPajamaGithub


text:
compiled with [ghc](https://www.haskell.org/downloads#platform)
```
$ ghc -o <name> --make <name>.hs
$ ghc -o add --make add.hs
```

parse.hs uses the Parsec library so you might need to build it like so

```
$ ghc -package parsec -o parse --make parse.hs
```

You can then run them like so

```
$ ./add 50 5000
5050
```


Approaching Red Sculpture - Walters Group Inc.
Two massive curved plates of steel finished in stunning red automotive paint, viewable from the ground or from a walkway connecting two towers in central Toronto. Conceived by artist Maha Mustafa and executed by Feature Walters, this public art instantly became a flagship piece in the already impressive holdings of Concord CityPlace.
Sometimes attaining visual simplicity requires complex planning, innovative thinking, and borderline obsessive behaviour.
In this case, the vision of artist Maha Mustafa was a piece of public art comprising two gracefully curved sheets of steel, painted bright shiny red, to be installed as part of a condominium tower complex in Toronto.
It is our dedication to always finding a way that put this one over the top and got it where the artist intended it to be. Rolling the steel was just the beginning. We spent days priming, sanding and wet sanding. The workers shone high-wattage illumination against the steel and they would rest their faces on the curves looking for irregularities and shadows. They were up on ladders and crawling all over the piece, getting it perfect. Any wrinkle or dimple would impact the high-gloss face and make it less than it should be. We knew that for this art to be effective, it had to be this flawless.
The work was taxing, but it paid off: The sculpture now sits outside two soaring Toronto towers, below a connecting sky bridge that also happens to be a Walters Group project. From the ground and from the sky, Approaching Red is an emblem of our work preserving the intent of the artist.
It was no simple job to turn this vision into reality. The artist was seeking a conical curve (as opposed to cylindrical), which was essentially impossible with steel structures of that thickness and size. While the finished product looks pure and simple, that appearance was only achieved through complex planning and execution.


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text:
Home » Q&A: Bentley Systems' Claire Rutkowski on Remote Work, Burnout and the Hybrid Office
BusinessTechWorkforceDiversity and InclusionInformation technology
Q&A: Bentley Systems' Claire Rutkowski on Remote Work, Burnout and the Hybrid Office
In response to the emergence of the COVID-19 pandemic, in April 2020 Bentley Systems made its cloud-based infrastructure design and collaboration software ProjectWise 365 available for free, waiving subscriptions for five months. Bentley CIO Claire Rutkowski says that while the technology to enable remote work and hybrid offices already exists, firms will have to adjust their management styles to this new long-term scenario.
KEYWORDS Bentley Systems / Business Management / Hybrid Office / remote work / Work from Home
Rutkowski
As chief information officer and senior vice president at software vendor Bentley Systems, Claire Rutkowski spends a great deal of time talking to Chief Technology Officers and other tech-minded people in AEC firms. Earlier this summer she hosted a panel at the Environmental Financial Consulting Group's Technology Leadership conference on what technology for remote work means in the future, as some firms transition to bring some workers back to the office. Rutkowski took the time to chat with ENR Senior Technology Editor Jeff Rubenstone about how technology is enabling long-term remote working, and what managers must learn to run a hybrid office. The transcript has been lightly edited for clarity.
ENR: Here we are, more than a year into the disruption from COVID-19, and many companies still have employees working from home or are beginning to bring them back to the office. What has been your experience with this, both at Bentley Systems and among the industry firms you're talking to regularly?
Rutkowski: This is a different world now. We're coming out of the pandemic into a world where employees have very high expectations and are being very clear about the fact that they don't want to go into the office every day. They want to have a hybrid model.
A lot of people are coming out of this [phase of] the pandemic and reassessing their life priorities and what they're doing. There's a lot of competition for qualified talent and a lot of that qualified talent is asking, do you have remote or hybrid options?
I think that the majority of companies are thinking they're going to have to go hybrid in order to attract the right talent, but the other thing important to note is that—while we don't know what's going to happen with the Delta variant—in the meantime life has continued. Businesses have continued, deliverables have been met, and despite the kind of old school thinking of "I have to see you to know you're working," I think people have realized throughout the pandemic that, in many cases, productivity has gone up, not down.
ENR: Being present in the office was a key part of many office cultures before the pandemic, with managers walking cubicle rows to confirm everyone is getting work done.
Pre-pandemic, most people were in the office. The managers did go up and down the rows, looking over people's shoulders, sitting in on design reviews and seeing literally see how things were going. During the pandemic everybody was remote, so managers have pretty much I think figured out the skill of managing remote colleagues. Or, if they haven't figured it out, the evidence of that is now becoming very visible.
But I think they're actually in for a harder time than that, because pre-pandemic was a level playing field, with most people in the office. Most people out of the office, working from home, is also [a level playing field]. Now we're going to get into a time where 70% of people want to be in or out, a hybrid [arrangement]. And some are going to be all the way at home and some are going to be all the way in the office for most companies. And it's not a level playing field anymore.
So the manager is going to have to learn those skills. "So now I have to manage remote? I have to manage in person and I have to make it feel fair and the same for everyone." That's a whole different set of challenges. That's a whole new ball game that I think is going to require training. It's going to require a real concentration on different things that people haven't had to deal with.
They're going to have to be the ones that make sure that there isn't an A Team in the office and the B Team that is remote. They're going to have to make sure that when they have meetings, they don't just have all the "in the office" people that day—they're going to have to be careful how they keep the team together because it won't be together, but it won't be everyone apart and [working] remotely either. It will be in between.
ENR: Do you think there's some anxiety among remote workers about not responding fast enough to requests, that persistent suspicion that managers think everyone who comes into the office is by default more on the ball?
Exactly, and I think that's something that managers are really going to have to watch for. They're going to have to mediate relationships between the people on their teams to make sure they are alert for resentment or jealousy, or that somebody who has chosen to be remote is kind of gradually shutting down, feeling less included and becoming disengaged. Because if the people who do not come in the office start feeling disengaged from the people who do then then you're going to see a loss of productivity and you don't want that right? I mean, purely from a work perspective. Lack of engagement is a lack of productivity is a loss of profit margin and revenue.
So I think managers are going to have to continue that understanding and empathy, and so are the leaders of the companies that used to have everyone in the office and decided to go hybrid. They can't choose to go hybrid to attract talent but then resent it everyday. Because that's not really fair either.
ENR: There's also the issue of overworking and the reluctance to take time off that was already there before the pandemic in this industry. You have some work cultures where no one wants to be seen as absent or not totally engaged at all times, which gets extra stressful in extended work-from-home scenarios.
Last year and this year, I have talked about and emailed and written about mental health to the people on my team more than I ever have before. Whether it was at the beginning of the pandemic or later on, I could tell that people were online at like crazy hours and they sounded kind of tired. And I know there's the stress of the pandemic and everything around it to take into account as well.
And that may have been some of it, but I think it was also was they were really afraid that they were going to be let go. And so I found myself in the kind of unusual position of having a mandatory "all hands" in which I told everybody to stop working so hard.
I've never done that before, but I knew they were working unhealthy hours and I had to explicitly say "look, no one is looking over your virtual shoulder." I get it, there's stuff happening, so let's find a way to buffer between work and home, and I gave them some suggestions and then it was an email all about self care. It's OK to not be OK. It's OK to cancel your meetings for the day. if you're just feeling anxious and stressed, you don't have to turn on the dang camera, you know? Because I know some people find that mentally exhausting.
ENR: Do you see similar signs of anxiety across the broader industry in your conversations with CTOs and others at firms?
I think people in general are getting burned out. Not only do they have the pressure of trying to keep up the momentum and keep up with deadlines, but they're doing it remotely and having to do virtual design reviews. Maybe they're using virtual reality if they're lucky enough, but otherwise everything is just a little bit harder to do. It's that overall uncertainty of just not knowing what's going to happen next [with COVID-19]. It's exhausting.
ENR: Do you get the sense that, over a year into this, companies and management are beginning to adapt to the new state of affairs?
On the bright side, I do think companies have realized through the pandemic that their colleagues can be trusted to be productive remotely, so they're more open to that now. They've had to leverage a lot of digital collaboration tools and get used to asynchronous communication.
I think companies are much better equipped to handle it and have the right tools at least to handle it. And so the emphasis, as it did for me, moves from tools and skills and equipping people to care about and support people, mental health, engagement, and inclusivity, particularly from middle managers, to level that playing field. I would like to think that makes the AEC industry a more inclusive and better place to work overall.
ENR: Construction is often seen as being behind the curve on new technology and management styles. Do you think the industry is going to adapt to these new ways of working?
For the most part people are going to embrace the opportunity here. I don't think of this as just a choice between mandating return to the office or enabling a hybrid working model or allowing people to work fully remote. It's a leadership challenge. But it's also a leadership opportunity, an opportunity for the chief executive
CEOs will say "I've seen you all work productively for the last year and a half. I trust you to make the right choices for you and for your families about where you want to spend your time and how you want to work. But I also trust that you'll focus and work productively at work—whether you're at work at your desk, or whether you're at work at home."
I think that CEOs who do this will find attracting and retaining talent easier, that the engagement and morale of their employees will improve and they will provide a much more inclusive workplace. I think they'll feel much more supported and welcomed to stay as long as they can work out the right kind of schedule with their manager.
For me it's a "lean in" moment. Are you going to do it or are you going to slip backwards?
Claire Rutkowski joined Bentley in 2016 from MWH Global where she served as chief information officer, responsible for delivering IT strategy, services, and support to 7,000 engineering professionals around the world. She now heads Bentley's global IT organization, shaping and delivering a technology agenda across its global business.
Bentley Systems Partners with Microsoft to Boost Cloud-based Digital Twin Capability
Agile Telepresence Robot Offers a Different Spin on Remote Meetings
Bentley Builds Out Its Offerings, With an Eye on the Full Construction Lifecycle

redpajama_set_name: RedPajamaCommonCrawl


Press the square Power Button in the center of the controller to turn it on.
Use the Arrow buttons to highlight the Driver Control program's icon (highlighted in the image).
Then press the A button to choose Drive.
Press the A button again to choose Run.
Press the B button to return to the previous menu, if needed.
The rectangular L1 button on the front of the controller closes the claw.
The rectangular L2 button on the front of the controller opens the claw.


!!Attention!! explor8ion.com is being updated and trip reports migrated to a new site while this one is still operational. The new version of this trip report can be found at https://verndewit.com/2010/09/04/perspective-in-an-overshared-world/ and contains more photos in a modern format. For more information on this move and possible future changes please click here.
Really well written Vern. I've read tons of your TR's but only just read your page/blog. This one in particular rings true with me. Gained a lot more respect for you.
Enjoyed the read, would you ever publish a manifesto of sorts? In the same vein as Royal Robbins but geared towards peak bagging?
Hey David, thanks for reading - I'm never sure if anyone bothers with reading my occasional rants. I have thought about cleaning this article up a bit and making it more visible but I want to be careful that I'm not just coming across as a "grumpy old guy". It's important to share perspectives without stomping over other people's freedoms or different ways of living their lives. I never know exactly where that line is - it can be a bit tricky.


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Early Origins of the Gulde family
The surname Gulde was first found in Suffolk where Hugo filius Goldce was listed in the Domesday Book of 1086. [4] "Golde and Goldus are personal names in the Domesday Book" [2]
Other early records include: Ralph filius Golde registered in the Pipe Rolls for Bedfordshire in 1193; Golde Bassat in the Hundredorum Rolls of Cheshire in 1279; Walter Golde in the Pipe Rolls for Devon in 1165; Ralph Golde in the Assize Rolls for Somerset in 1268; and John Gulde in Cornwall in 1297. [1]
The Hundredorum Rolls has the name listed as a forename and a surname: Adam Gold; Golda Imayn; and Golda Bassat, Cambridgeshire. [2]
The Yorkshire Poll Tax Rolls includes Willelmus Golde, 1370; and Ricardus Gold, 1379.
In Scotland, more often than not, the name is spelt Guild and variants of the same. It is generally thought that Goold (Gold) is a variant. [5] [3]
Here, the first record was of Alexander Gulde who rendered to Exchequer the accounts of the bailies of Stirling, 1421 and 1425. Thomas Gulde and William Gulid, 'husbandis,' were charter witnesses, 1481, and Henry Guld is mentioned in a Scone document of 1491. [5]
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Early History of the Gulde family
This web page shows only a small excerpt of our Gulde research. Another 195 words (14 lines of text) covering the years 1086, 1193, 1312, 1500, 1502, 1512, 1521, 1584, 1593, 1676, 1659, 1676, 1619, 1675, 1644, 1710, 1710, 1794, 1709, 1586, 1657, 1586, 1657, 1734, 1678 and are included under the topic Early Gulde History in all our PDF Extended History products and printed products wherever possible.
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Gulde Spelling Variations
The English language only became standardized in the last few centuries; therefore, spelling variations are common among early Anglo-Saxon names. As the form of the English language changed, even the spelling of literate people's names evolved. Gulde has been recorded under many different variations, including Gould, Goold, Gold, Gilder and others.
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Early Notables of the Gulde family (pre 1700)
Notables of the family at this time include James Gould (1593-1676), an English politician who sat in the House of Commons between 1659 and 1676; and Thomas Gould (ca. 1619-1675), the first pastor of the First Baptist Church of Boston in Boston, Massachusetts. Sir Henry Gould the Elder (1644-1710), was an English judge, son and heir of Andrew Gould of Winsham, Somersetshire. Sir Henry Gould the Younger (1710-1794), was an English judge, the fourth son of Davidge Gould of Sharpham...
Another 78 words (6 lines of text) are included under the topic Early Gulde Notables in all our PDF Extended History products and printed products wherever possible.
Migration of the Gulde family to Ireland
Some of the Gulde family moved to Ireland, but this topic is not covered in this excerpt.
Another 116 words (8 lines of text) about their life in Ireland is included in all our PDF Extended History products and printed products wherever possible.
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Gulde migration to the United States +
For many English families, the political and religious disarray that shrouded England made the far away New World an attractive prospect. On cramped disease-ridden ships, thousands migrated to those British colonies that would eventually become Canada and the United States. Those hardy settlers that survived the journey often went on to make important contributions to the emerging nations in which they landed. Analysis of immigration records indicates that some of the first North American immigrants bore the name Gulde or a variant listed above:
Gulde Settlers in United States in the 18th Century
Rolle Gulde, who arrived in Pennsylvania in 1749 [6]
Contemporary Notables of the name Gulde (post 1700) +
William O. Gulde, American politician, Mayor of Hempstead, New York, 1955-65 [7]
Related Stories +
Family Crests: the Heraldic Artist
Family Crests: Timeline
Family Crests: Elements
Family Crests and Genealogy: how they relate
Spelling variations
Hundredorum Rolls
Assize Rolls
Pipe Rolls
Filius
Citations +
^ Reaney, P.H and R.M. Wilson, A Dictionary of English Surnames. London: Routledge, 1991. Print. (ISBN 0-415-05737-X)
^ Bardsley, C.W, A Dictionary of English and Welsh Surnames: With Special American Instances. Wiltshire: Heraldry Today, 1901. Print. (ISBN 0-900455-44-6)
^ Smith, Eldson Coles, New Dictionary of American Family Names New York: Harper & Row, 1956. Print
^ Williams, Dr Ann. And G.H. Martin, Eds., Domesday Book A Complete Translation. London: Penguin, 1992. Print. (ISBN 0-141-00523-8)
^ Black, George F., The Surnames of Scotland Their Origin, Meaning and History. New York: New York Public Library, 1946. Print. (ISBN 0-87104-172-3)
^ Filby, P. William, Meyer, Mary K., Passenger and immigration lists index : a guide to published arrival records of about 500,000 passengers who came to the United States and Canada in the seventeenth, eighteenth, and nineteenth centuries. 1982-1985 Cumulated Supplements in Four Volumes Detroit, Mich. : Gale Research Co., 1985, Print (ISBN 0-8103-1795-8)
^ The Political Graveyard: Alphabetical Name Index. (Retrieved 2016, January 20) . Retrieved from http://politicalgraveyard.com/alpha/index.html
Gulde (English)
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text:
Refugees face violence, abuse on Europe's new Balkan route
April 16, 2019 Deutsche Welle
It has become increasingly difficult for refugees to make their way from Greece to Central Europe. But in the face of police violence and the risk of arrest, thousands still embark on the dangerous journey.
In early April, Greek police near the northern city of Thessaloniki used tear gas to prevent refugees from making their way to the border with North Macedonia. The refugees were headed to Central and Western Europe, driven by a series of false reports on social media that the border crossings to North Macedonia and Albania had been opened. Around 2000 people believed the rumor and headed for northern Greece. After three days, the last of them gave up and returned to refugee shelters.
Many of the roughly 70 000 refugees in Greece and the Western Balkans want to travel further north, but numerous obstacles stand in their way. North Macedonia and Hungary erected barbed wire fences on their southern borders years ago. Meanwhile, Bulgaria is known for its harsh treatment of refugees. Human rights organisations and the European Commission have accused the country of violating European Union asylum laws.
To continue reading, please visit dw.de
(Archive photo: rescue.org.uk)
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May 16, 2022 May 16, 2022 The Sofia Globe staff
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text:
Michael Zinni (September 10, 1948 – July 29, 2014) was the head coach of the Minnesota State University, Mankato golf team from 2003 to 2009.

Zinni also played on the Champions Tour for senior golfers.  He was the head golf professional at the Mankato Golf Club. Zinni died on July 29, 2014 at Mayo Clinic from a lung infection.

References

External links

American male golfers
PGA Tour Champions golfers
College golf coaches in the United States
Golfers from Minnesota
Minnesota State University Moorhead
1948 births
2014 deaths

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