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Spectral cotangent space

I’m supposed to give a 10 minute talk on what I’ve been up to recently. 10 minute talks are pretty brutally short, though I think they’re better than 20 minute talks — those give you the illusion that you might be able to convey something, but you really just flatly can’t. With only 10 minutes you’re not kidding anyone.

Even so, I should still practice, and writing out what I want to say is a good place to start, and in turn a good blog post.

So, in chromatic homotopy theory a basic object of study is $E$-theory, which is a functor $\mathsf{Spectra} \to \mathsf{QCoh}(LT_n, \Gamma)$ from the category of spectra to the category of quasicoherent sheaves on the $n$th Lubin-Tate deformation space, equivariant against the action of the Morava stabilizer group $\Gamma$. This is some fancied up way of describing a homology theory $E_n^\vee$, taking values in $(E_n^\vee, E_n^\vee E_n)$-comodules. This functor factors through a localization factor to the category of $K(n)$-local spectra, and the whole of chromatic homotopy theory revolves around trying to understand $E$-theory and this localized category.

In the parent category of spectra, we are especially interested in spheres, and one characterization of spheres is that they are “invertible” with respect to the monoidal structure endowed by the smash product, in the sense that for each $n$ we have $\mathbb{S}^n \wedge \mathbb{S}^{-n} \simeq \mathbb{S}^0$, and this happens for no other spectra. The smash product descends to the $K(n)$-local category, where we can consider spectra $X$ with the same property: there is some $Y$ with $X \wedge Y = L_{K(n)} \mathbb{S}^0$. This group, denoted $\mathrm{Pic}_n$, turns out to be much more interesting — it is a p-adic analytic group. Understanding this group has turned out to be related to understanding the homotopy groups of the $K(n)$-local sphere, and so $\mathrm{Pic}_n$ is both hard to describe and greatly interesting.

There are also invertible objects in the target category of the $E$-theory functor: these are the equivariant line bundles. The Kunneth theorem shows that, in fact, the $E$-theory functor restricted to $\mathrm{Pic}_n$ factors through the subcategory of line bundles. We can also take the pullback of the corner whose edges are $K(n)$-local spectra and equivariant line bundles mapping to / including into equivariant sheaves on Lubin-Tate space — these are of course the $K(n)$-local spectra whose image under the E-theory functor is one of these line bundles. Hopkins, Mahowald, and Sadofsky show that the induced map from $\mathrm{Pic}_n$ to this pullback is an isomorphism (maybe this was known before them, but it seems to first appear there). This is a very useful detection theorem for when we’ve found an element of the Picard group, and so one thing we can do is to try to produce examples of line bundles first, and then lift these examples to spectra.

So, we want to build 1-dimensional free modules, essentially. One way you can get such a thing is to take a 1-dimensional smooth variety and build its cotangent sheaf — this gives a map $T^*$ from such things to 1-dimensional free modules. We can additionally suppose that we have some spectrum $X$ in mind so that $\mathrm{Spf}\;E_n^* X$ is such a 1-dimensional smooth formal variety. (An important thing here is that $X$ needs a diagonal map so that cohomology is ring-valued.) What we’re looking for, then, is a kind of interchange law — starting with X we can apply $\mathrm{Spf}\;E_n^*$ to get a variety and then $T^*$ to get the module, but we would like to have a functor $\tilde T^*$ which is spectrum-valued, and then $E_n^\vee \tilde T^* X = T^* \mathrm{Spf}\; E_n^* X$. This $\tilde T^*$ is what I’ve constructed.

So, with such a $\tilde T^*$, one can ask what good it is — we need to get some interesting things out, or we haven’t really accomplished our goal of learning more about $\mathrm{Pic}_n$. We’ll look at $\tilde T^* \Sigma^\infty_+ X$ for some input spaces $X$. When $X = \mathbb{C}\mathrm{P}^\infty$, it turns out we get $\mathbb{S}^2$. When $X = \mathbb{R}\mathrm{P}^\infty$, we get $\mathbb{S}^1$. The first interesting example comes from a calculation of Ravenel and Wilson (yes, that thing again): $X = \underline{\smash{H\mathbb{Z}/p^\infty}}_n$ has this property as well. In that case, it turns out that we get a spectrum whose image is the determinantally twisted line bundle, and so this is sometimes called the determinantal sphere.

The determinantal sphere appears in a few places — it’s really the only important example of an element of the $K(n)$-local Picard group that we know for all n. For instance, the Brown-Comenetz dual of the sphere, investigated by Hopkins and Gross, turns out to have this same property. So does a spectrum constructed by Goerss, Henn, Mahowald, and Rezk, in their paper on $\mathrm{Pic}_2$ at $p = 3$, though there’s some buzz about these two examples differing by a smash factor invisible to the E-theory functor (called an “exotic element”). There are also a lot of structural things that this construction points out! It’s all very interesting. But, in 10 minutes, that’s probably as much as I can hope to say — so that’s where I’ll quit.


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Determinants – Part 3

Hello everyone, I hope you are all surviving your semester, wherever you may be! In this post, I want to show that (non-zero) alternating n-linear forms always exist over any vector space of dimension n. We already know that, if a non-zero alternating n-linear form $\omega$ on $V$ exists, then any other alternating n-linear form must be a scalar multiple of $\omega$. We’ll use this fact, in a somewhat roundabout way, to define the determinant of a linear transformation. Then we can show the connections between determinants and matrices. The proof of the existence statement uses facts about the dual space $V'$ of $V$, so I want to review some basic facts about this space first.

Definitions: If $V$ is a vector space (finite or infinite dimensional) over a scalar field $\mathcal{F}$, a linear operator $\phi:V\mapsto\mathcal{F}$ is called linear functional on $V$. The vector space of all linear functionals on $V$ is called the dual space of $V$ and is denoted by $V'$. The dual space $(V')'$ of $V'$ is often called the double dual of $V$ and is often denoted $V''$.

The reader should verify that indeed $V'$ is a vector space under the usual definitions of function addition and multiplication. For purposes of this post, we will only care about finite dimensional vector spaces, but it is interesting to note that for an infinite dimensional space, the (algebraic) dual is much larger than $V$. For this reason, functional analysts often restrict themselves to working with continuous linear functionals on $V$. In the future I hope to write about some of the differences and similarities between “algebraic” linear algebra and “analytic” linear algebra. Many of the results for infinite dimensional spaces (in functional analysis) are similar to results for finite dimensional linear algebra (algebraic results), however new proofs need to be introduced to work in analytic notions of continuity and convergence. This is a beautiful area of mathematics that I hope to cover eventually. For now though, we have the following:

Proposition: If $V$ is finite dimensional, say of dimension n, then $V'$ also has dimension n. In particular these spaces are isomorphic.
Proof: That $V\simeq V'$ follows immediately from the fact that the two spaces have the same finite dimension, so it suffices to prove the first statement. Recall that any linear transformation $L$ on $V$ can be defined by its action on a basis $\{x_i\}_{i=1}^n$ of $V$ (to see this, given $x\in V$, write $x=\sum_i c_ix_i$, where this expansion is unique. Then by linearity, $Lx=\sum_i c_iLx_i$). Given such a basis $\{x_i\}$, define $y_i\in V'$ by $y_i(x_j)=\delta_{i,j}$, so $y_i(x_j)=1$ if $i=j$, otherwise $y_i(x_j)=0$. I claim $\{y_i\}_{i=1}^n$ is a basis for $V'$. To this end, suppose $y=\sum_i \alpha_iy_i=0$. Then for each $x\in V$ we have $y(x)=0$, so in particular $0=y(x_j)=\sum_i \alpha_iy_j(x_j)=\alpha_j$. Thus the $y_i$ are linearly independent. Moreover, given any $y\in V$, write $\eta_i=y(x_i)$. Then I claim $y=\sum_i\eta_iy_i$. To see this, given $x\in V$, write $x=\sum_i\alpha_i x_i$. Then $y(x)=\sum_i \alpha_i\eta_i$, and for each $j$ we have $\eta_jy_j(x)=\eta_j\alpha_j$, so summing over $j$ gives the result.

This result shows then that for finite dimensional spaces $V''\simeq V$ as well. Now, back to our discussion of alternating forms: why do we care about linear functionals? Well, for one thing, a linear functional is an alternating 1-linear form on $V$, so this fact shows there are non-trivial 1-linear functionals on any non-trivial vector space, a fact that we will use in the next proof. Moreover, we will use these notions to define our larger n-linear form on $V$. Dual spaces are very interesting, but they are not the main focus of this post. Let’s now turn to the main theorem in this post, and probably the most important theorem in this series on determinants:

Theorem Let $V$ be an n-dimensional vector space. Then there is a non-zero alternating n-linear form on $\displaystyle\bigoplus_{i=1}^nV$.
Proof: I’ll show that for any $1\leq k\leq n$ that there is an alternating k-linear form on $V$. If $k=1$, then the result holds because the dual space of $V$ is non-trivial. Now let $1\leq k and suppose that $\omega$ is a non-zero alternating k-linear form. Thus there are $\{z_i\}_{i=1}^k\subset V$ with $\omega(z_1,\dots,z_k)\neq0$. Choose a vector $z_{k+1}\notin\text{span}\{z_1,\dots,z_k\}$ (such a vector exists because $k). There is a linear functional $\phi\in V'$ such that $\phi(z_i)=0$ for $1\leq i\leq k$ and $\phi(z_{k+1})\neq 0$ (for instance we could choose $\phi$ to be the standard dual basis vector since $z_1,\dots,z_k$ must be linearly independent by a proposition in the previous post). Define a function $\nu$ by

$\dagger\hspace{3mm} \displaystyle\nu(x_1,\dots,x_k,x_{k+1})=\left(\sum_{i=1}^k(i,k+1)(\omega(x_1,\dots,x_k)\phi(x_{k+1}))\right)-\omega(x_1,\dots,x_k)\phi(x_{k+1})$

so, for instance, if $k=3$, then

$\nu(x_1,x_2,x_3,x_4)=\omega(x_4,x_2,x_3)\phi(x_1)+\omega(x_1,x_4,x_3)\phi(x_2)+\dots-\omega(x_1,x_2,x_3)\phi(x_4)$

The reader may verify that $\nu$ is indeed a k+1-linear form on $V$. Moreover, $V$ is non-zero, for we have $\nu(z_1,\dots,z_k,z_{k+1})=-\omega(z_1,\dots,z_k)\nu(z_{k+1})\neq0$ by construction (since $\nu(z_j)=0$ for $1\leq j\leq k$), and so it remains to show $\nu$ is alternating. To that end suppose $x_1,\dots,x_{k+1}\subset V$ with $x_j=x_i$ for some $1\leq i< j\leq k+1$. Now, because $\omega$ is alternating, and $x_i,x_j$ occur in the argument of $\omega$ in all but two terms in the sum on the right in $\dagger$, the sum on the right collapses to a sum of two terms. There are two cases: if $j=k+1$, then the sum is

$\nu(x_1,\dots,x_{k+1})=(i,k+1)\omega(x_1,\dots,x_k)\phi(x_{k+1})-\omega(x_1,\dots,x_k)\phi(x_{k+1})$

which is clearly zero because $x_i=x_{k+1}$. Otherwise, if $1\leq i, then write $x=x_i=x_j$ so that the terms in the sum will look like

$\omega(x_1,\cdots,x_{k+1},\cdots,x,\cdots,x_k)\phi(x)+\omega(x_1,\cdots,x_{k+1},\cdots ,x,\cdots, x_k)\phi(x)$

and from this we can see that if we apply the permutation $(i,j)$ to the $\omega$ term on the right, then the arguments in $\omega$ agree, so since it is alternating they differ by only a sign (recall this fact from a previous post), and hence the sum is indeed zero. This shows that $\nu$ is alternating, which completes the proof.

So, that was messy! What did we do? Well, we start by finding a non-zero linear functional, and associate with it some subspace of $V$. Then we pick something in the annihilator of that subspace (and a corresponding vector in $V$) and build out of it an alternating bi-linear form. We can be sure that it is non-zero because we set up our subspaces of $V$ so that this function is non-zero when vectors are chosen correctly. Moreover, to evaluate the function, we hold a lot of things fixed and permute two different entries and these permutation constructions, combined with the minus sign in the construction of $\nu$, ensure that this new form is alternating. It’s not a very insightful proof, but you can certainly see the connections to determinants of matrices where we fix one row (which corresponds to $\phi$) and vary the arguments $\omega$ (which corresponds to taking determinants of the sub-matrices). Now, at long last, we are ready to define the determinant of a linear transformation. After that, in the following post, I will connect all these notions to matrices and prove that the matrix determinant is in fact an alternating n-linear form when thought of as a function on rows.

So how do we construct the determinant? Recall that any linear transformation $L$ on a 1 dimensional space must be multiplication by some scalar $c_L$. To see this, note that in a one dimensional space, every vector $v=cx$, for some vector $x$, thus $Lx=c_Lx$, and for any vector $y$ we have $Ly=L(\alpha x)=\alpha Lx=\alpha c_L x=c_L\alpha x=c_Ly$. Now let $A$ be a linear operator on $V$. Define a map $\overline{A}$ on the space of alternating n-linear forms of $V$ by $[\overline{A}(\omega)](x_1,\dots,x_n)=\omega(Ax_1,\dots,Ax_n)$. Then the reader may verify that $\overline{A}$ is a linear operator. Since the space of n-linear forms is one dimensional, there is a unique scalar $\delta_A$ so that $\overline{A}(\omega)=\delta_A\omega$ for every alternating n-linear form $\omega$ on $V$. This unique scalar $\delta_A$ is called the determinant of the linear transformation $A$ and is denoted by $det(A)$. For fun, let’s compute a few values of $det$. Suppose $Ax=\alpha x$ for all $x\in V$, i.e. $A$ is multiplication by some fixed scalar. Then

$[\overline{A}(\omega)](x_1,\dots,x_n)=\omega(\alpha x_1,\dots,\alpha x_n)=\alpha^n\omega(x_1,\dots,x_n)$

and so we have $det(A)=\alpha^n$. In particular then $det(I)=1$ and $det(0)=0$. Finally, if $A,B$ are both linear transformations, let $C=AB$ (the composition) and note that

$[\overline{C}(\omega)](x_1,\dots,x_n)=\omega(ABx_1,\dots,ABx_n)=[\overline{A}(\omega)](Bx_1,\dots,Bx_n)=[\overline{B}\overline{A}(\omega)](x_1,\dots,x_n)$

and thus $\overline{C}=\overline{B}\overline{A}$. But then for any alternating n-linear form $\omega$ on $V$ we have

$det(B)det(A)\omega=det(B)(\overline{A}(\omega))=\overline{B}\overline{A}(\omega)=\overline{C}(\omega)=det(C)\omega$

from which it follows that $det(AB)=det(A)det(B)$, since scalars commute. Okay! That was a lot for one post! Now that we have defined the determinant, all we have left to do is connect it to matrices and then I’ll show you some cool theorems involving determinants and volumes of sets in $\mathbb{R}^n$. For now though, you’ll have to be content with just this. Enjoy!

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Scalar boundary value problems on junctions of thin rods and plates
ESAIM: Mathematical Modelling and Numerical Analysis - Modélisation Mathématique et Analyse Numérique, Tome 48 (2014) no. 5, p. 1495-1528
Le texte intégral des articles récents est réservé aux abonnés de la revue. Consulter l'article sur le site de la revue
We derive asymptotic formulas for the solutions of the mixed boundary value problem for the Poisson equation on the union of a thin cylindrical plate and several thin cylindrical rods. One of the ends of each rod is set into a hole in the plate and the other one is supplied with the Dirichlet condition. The Neumann conditions are imposed on the whole remaining part of the boundary. Elements of the junction are assumed to have contrasting properties so that the small parameter, i.e. the relative thickness, appears in the differential equation, too, while the asymptotic structures crucially depend on the contrastness ratio. Asymptotic error estimates are derived in anisotropic weighted Sobolev norms.
DOI : https://doi.org/10.1051/m2an/2014007
Classification:  35B40,  35C20,  74K30
@article{M2AN_2014__48_5_1495_0,
author = {Bunoiu, R. and Cardone, G. and Nazarov, S. A.},
title = {Scalar boundary value problems on junctions of thin rods and plates},
journal = {ESAIM: Mathematical Modelling and Numerical Analysis - Mod\'elisation Math\'ematique et Analyse Num\'erique},
publisher = {EDP-Sciences},
volume = {48},
number = {5},
year = {2014},
pages = {1495-1528},
doi = {10.1051/m2an/2014007},
mrnumber = {3264363},
language = {en},
url = {http://www.numdam.org/item/M2AN_2014__48_5_1495_0}
}

Bunoiu, R.; Cardone, G.; Nazarov, S. A. Scalar boundary value problems on junctions of thin rods and plates. ESAIM: Mathematical Modelling and Numerical Analysis - Modélisation Mathématique et Analyse Numérique, Tome 48 (2014) no. 5, pp. 1495-1528. doi : 10.1051/m2an/2014007. http://www.numdam.org/item/M2AN_2014__48_5_1495_0/

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[51] S.A. Nazarov, Asymptotic expansions at infinity of solutions of a problem in the theory of elasticity in a layer. Tr. Mosk. Mat. Obs. 60 (1999) 3-97; Trans. Moscow Math. Soc. (1999) 1-85. | MR 1702684 | Zbl 0944.74011

[52] S.A. Nazarov, Junctions of singularly degenerating domains with different limit dimensions. II. Tr. Semin. im. I. G. Petrovskogo 20 (2000) 155-195; 312-313; J. Math. Sci. 97 (1999) 155-195. | MR 1846015 | Zbl 0976.35013

[53] S.A. Nazarov, Asymptotic analysis and modeling of the junction of a massive body and thin rods. Tr. Semin. im. I. G. Petrovskogo 24 (2004) 95-214, 342-343; J. Math. Sci. 127 (2005) 2192-2262. | MR 2360841 | Zbl 1145.74390

[54] S.A. Nazarov, Estimates for the accuracy of modeling boundary value problems on the junction of domains with different limit dimensions. Izv. Ross. Akad. Nauk Ser. Mat. 68 (2004) 119-156; Izv. Math. 68 (2004) 1179-1215. | MR 2108526 | Zbl 1167.35343

[55] S.A. Nazarov, Elliptic boundary value problems on hybrid domains. Funktsional. Anal. i Prilozhen 38 (2004) 55-72; Funct. Anal. Appl. 38 (2004) 283-297. | MR 2117508 | Zbl 1127.35328

[56] S.A. Nazarov, Korn's inequalities for elastic joints of massive bodies, thin plates, and rods. Uspekhi Mat. Nauk 63 (2008) 379, 37-110; Russian Math. Surveys 63 (2008) 35-107. | MR 2406182 | Zbl 1155.74027

[57] S.A. Nazarov, Asymptotic behavior of the solutions of the spectral problem of the theory of elasticity for a three-dimensional body with a thin coupler. Sibirsk. Mat. Zh. 53 (2012) 345-364; Sib. Math. J. 53 (2012) 274-290. | MR 2975940 | Zbl pre06105679

[58] S.A. Nazarov and B.A. Plamenevsky, Elliptic problems in domains with piecewise smooth boundaries. Moscow: Nauka. (1991); de Gruyter Expositions Math., vol. 13. Walter de Gruyter & Co., Berlin (1994). | MR 1283387 | Zbl 0806.35001

[59] G.P. Panasenko, Multi-scale Modeling for Structures and Composites. Springer, Dordrecht (2005). | MR 2133084

[60] G. Pólya and G. Szegö, Isoperimetric Inequalities in Mathematical Physics, Annal. Math. Studies, vol. 27, Princeton University Press, Princeton (1951). | Zbl 0044.38301

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text:
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## Section2.2Definition of a group

### Subsection2.2.1

We will use the notation $\ast \colon S\times S\to S$ to denote a binary operation on a set $S$ that sends the pair $(x,y)$ to $x\ast y\text{.}$ Recall that a binary operation $\ast$ is associative means that $x\ast(y\ast z)= (x\ast y)\ast z$ for all $x,y,z\in S\text{.}$

###### Definition2.2.1.Group.

A group is a set $G\text{,}$ together with a binary operation $\ast\colon G\times G \to G$ with the following properties.

• The operation $\ast$ is associative.
• There exists an element $e$ in $G\text{,}$ called an identity element, such that $e\ast g=g\ast e=g$ for all $g\in G\text{.}$
• For every $g\in G\text{,}$ there exists an element $h\in G\text{,}$ called an inverse element for $g\text{,}$ such that $g\ast h=h \ast g=e\text{.}$
###### Definition2.2.3.Multiplicative notation.

Let $G$ be a group. By Proposition 2.2.2, we may speak of an identity element as the identity element for $G\text{.}$ Given $g\in G\text{,}$ we may refer to an inverse element for $g$ as the inverse of $g\text{,}$ and we write $g^{-1}$ to denote this element. In practice, we often omit the operator $\ast\text{,}$ and simply write $gh$ to denote $g\ast h\text{.}$ We adopt the convention that $g^0$ is the identity element. For $k\geq 1\text{,}$ we write $g^k$ to denote $\underbrace{g\ast g\ast \cdots \ast g}_{k \text{ factors}}$ and we write $g^{-k}$ to denote $\left(g^{k}\right)^{-1}\text{.}$ This set of notational conventions is called multiplicative notation .

###### Definition2.2.4.Abelian group, additive notation.

In general, group operations are not commutative. 1  A group with a commutative operation is called Abelian.

For some Abelian groups, such as the group of integers, the group operation is called addition, and we write $a+b$ instead of using the multiplicative notation $a\ast b\text{.}$ We write $0$ to denote the identity element, we write $-a$ to denote the inverse of $a\text{,}$ and we write $ka$ to denote $\underbrace{a+ a+ \cdots +a}_{k \text{ summands}}$ for positive integers $k\text{.}$ This set of notational conventions is called additive notation .

Recall that a binary operation $\ast$ on a set $S$ is called commutative if $x \ast y = y\ast x$ for all $x,y\in S\text{.}$
###### Definition2.2.5.Order of a group.
The number of elements in a finite group is called the order of the group. A group with infinitely many elements is said to be of infinite order. We write $|G|$ to denote the order of the group $G\text{.}$
###### Definition2.2.6.The trivial group.

A group with a single element (which is necessarily the identity element) is called a trivial group. In multiplicative notation, one might write $\{1\}\text{,}$ and in additive notation, one might write $\{0\}\text{,}$ to denote a trivial group.

### Exercises2.2.2Exercises

###### 1.Uniqueness of the identity element.

Let $G$ be a group. Suppose that $e,e'$ both satisfy the second property of the Definition 2.2.1, that is, suppose $e\ast x=x\ast e = e'\ast x=x\ast e'=x$ for all $x\in G\text{.}$ Show that $e=e'\text{.}$

###### 2.Uniqueness of inverse elements.

Let $G$ be a group with identity element $e\text{.}$ Let $g\in G$ and suppose that $g\ast h = h\ast g = g\ast h' = h'\ast g = e\text{.}$ Show that $h=h'\text{.}$

###### 3.The cancellation law.

Suppose that $gx=hx$ for some elements $g,h,x$ in a group $G\text{.}$ Show that $g=h\text{.}$ [Note that the same proof, mutatis mutandis, shows that if $xg=xh\text{,}$ then $g=h\text{.}$]

###### 4.The "socks and shoes" property.

Let $g,h$ be elements of a group $G\text{.}$ Show that $(gh)^{-1} = h^{-1}g^{-1}\text{.}$

###### 5.Product Groups.

Given two groups $G,H$ with group operations $\ast_G,\ast_H\text{,}$ the Cartesian product $G\times H$ is a group with the operation $\ast_{G\times H}$ given by

\begin{equation*} (g,h)\ast_{G\times H} (g',h')= (g\ast_G g',h\ast_H h'). \end{equation*}

Show that this operation satisfies the definition of a group.

###### 6.Cyclic groups.

A group $G$ is called cyclic if there exists an element $g$ in $G\text{,}$ called a generator, such that the sequence

\begin{equation*} \left(g^k\right)_{k\in \Z}=(\ldots,g^{-3},g^{-2},g^{-1},g^0,g^1,g^2,g^3,\ldots) \end{equation*}

contains all of the elements in $G\text{.}$

1. The group of integers is cyclic. Find all of the generators.
2. The group $\Z_8$ is cyclic. Find all of the generators.
3. The group $\Z_2\times \Z_3$ is cyclic. Find all of the generators.
4. Show that the group $\Z_2\times \Z_2$ is not cyclic.
5. Let $m,n$ be positive integers. Show that the group $\Z_m\times \Z_n$ is cyclic if and only if $m,n$ are relatively prime, that is, if the greatest common divisor of $m,n$ is 1.
Hint
For the last part, observe that $(a,b)\in \Z_m\times \Z_n$ is a generator if and only if every entry in the sequence
\begin{equation*} (a,b),(2a,2b),(3a,3b),\ldots,(mna,mnb) \end{equation*}
is distinct (say why!). Let $L$ be the least common multiple of $n,m\text{.}$ If $m,n$ are relatively prime, then $L=mn\text{,}$ and if $m,n$ are not relatively prime, then $L\lt mn$ (say why!). Use this observation to prove the statement in the exercise.
###### 7.Cyclic permutations.

Let $n$ be a positive integer and $k$ be an integer in the range $1\leq k\leq n\text{.}$ A permutation $\pi\in S_n$ (see Definition 2.1.1) is called a $k$-cycle if there is a $k$-element set $A=\{a_1,a_2,\ldots,a_k\}\subseteq \{1,2,\ldots,n\}$ such that $\pi(a_i)=a_{i+1}$ for $1\leq i\leq k-1$ and $\pi(a_k)=a_1\text{,}$ and $\pi(j)=j$ for $j\not\in A\text{.}$ We use cycle notation $(a_1a_2\cdots a_k)$ to denote the $k$-cycle that acts as

\begin{equation*} a_1{\to} a_2{\to} a_3{\to} \cdots{\to} a_k{\to} a_1 \end{equation*}

on the distinct positive integers $a_1,a_2,\ldots,a_k\text{.}$ For example, the element $\pi=[1,4,2,3]=(2,4,3)$ is a 3-cycle in $S_4$ because $\pi$ acts on the set $A=\{2,3,4\}$ by

\begin{equation*} 2\to 4\to 3\to 2 \end{equation*}

and $\pi$ acts on $A^c=\{1\}$ as the identity. Note cycle notation is not unique. For example, we have $(2,4,3)=(4,3,2)=(3,2,4)$ in $S_4.$ Cycles of any length (any positive integer) are called cyclic permutations. A 2-cycle is called a transposition.

1. Find all of the cyclic permutations in $S_3\text{.}$ Find their inverses.
2. Find all of the cyclic permutations in $S_4\text{.}$
###### 8.

Cycles $(a_1a_2\cdots a_k)$ and $(b_1b_2\cdots b_\ell)$ are called disjoint if the the sets $\{a_1,a_2,\ldots,a_k\}$ and $\{b_1,b_2,\ldots,b_\ell\}$ are disjoint, that is, if $a_i\neq b_j$ for all $i,j\text{.}$ Show that every permutation in $S_n$ is a product of disjoint cycles.

###### 9.

Show that every permutation in $S_n$ can be written as a product of transpositions.

###### 10.Parity of a permutation.
1. Suppose that the identity permutation $e$ in $S_n$ is written as a product of transpositions
\begin{equation*} e=\tau_1\tau_2\cdots \tau_r. \end{equation*}
Show that $r$ is even.
2. Suppose that $\sigma$ in $S_n$ is written in two ways as a product of transpositions.
\begin{equation*} \sigma = (a_1b_1)(a_2b_2)\cdots (a_sb_s) = (c_1d_1)(c_2d_2)\cdots (c_td_t) \end{equation*}
Show that $s,t$ are either both even or both odd. The common evenness or oddness of $s,t$ is called the parity of the permutation $\sigma\text{.}$
3. Show that the parity of a $k$-cycle is even if $k$ is odd, and the parity of a $k$-cycle is odd if $k$ is even.
Hint
1. Consider the two rightmost transpositions $\tau_{r-1}\tau_{r}\text{.}$ They have one of the following forms, where $a,b,c,d$ are distinct.
\begin{equation*} (ab)(ab), (ac)(ab), (bc)(ab), (cd)(ab) \end{equation*}
The first allows you to reduce the transposition count by two by cancelling. The remaining three can be rewritten.
\begin{equation*} (ab)(bc), (ac)(cb), (ab)(cd) \end{equation*}
Notice that the index of the rightmost transposition in which the symbol $a$ occurs has been reduced by 1 (from $r$ to $r-1$). Finish this reasoning with an inductive argument.
###### 11.Cayley tables.

The Cayley table for a finite group $G$ is a two-dimensional array with rows and columns labeled by the elements of the group, and with entry $gh$ in position with row label $g$ and column label $h\text{.}$ Partial Cayley tables for $S_3$ (Figure 2.2.7) and $D_4$ (Figure 2.2.8) are given below.

1. Fill in the remaining entries in the Cayley tables for $S_3$ and $D_4\text{.}$
2. Prove that the Cayley table for any group is a Latin square. This means that every element of the group appears exactly once in each row and in each column.
Answer 1
\begin{equation*} \begin{array} {c|cccccc} \amp e \amp (23) \amp (13) \amp (12) \amp (123) \amp (132) \\ \hline e \amp e \amp (23) \amp (13) \amp (12) \amp (123) \amp (132) \\ (23) \amp (23) \amp e \amp (123) \amp (132) \amp (13) \amp (12)\\ (13) \amp (13) \amp (132) \amp e \amp (123) \amp (12) \amp (23)\\ (12) \amp (12) \amp (123) \amp (132) \amp e \amp (23) \amp (13)\\ (123) \amp (123) \amp (12) \amp (23) \amp (13) \amp (132) \amp e\\ (132) \amp (132) \amp (13) \amp (12) \amp (23) \amp e \amp (123) \end{array} \end{equation*}
Answer 2
\begin{equation*} \begin{array} {c|cccccccc} \amp F_V \amp F_H \amp F_D \amp F_{D'} \amp R_{1/4} \amp R_{1/2} \amp R_{3/4} \amp R_0\\ \hline F_V \amp R_0 \amp R_{1/2} \amp R_{3/4} \amp R_{1/4} \amp F_{D'} \amp F_H \amp F_D \amp F_V \\ F_H \amp R_{1/2} \amp R_0 \amp R_{1/4} \amp R_{3/4} \amp F_D \amp F_V \amp F_{D'} \amp F_H \\ F_D \amp R_{1/4} \amp R_{3/4} \amp R_0 \amp R_{1/2} \amp F_V \amp F_{D'} \amp F_H \amp F_D \\ F_{D'} \amp R_{3/4} \amp R_{1/4} \amp R_{1/2} \amp R_0 \amp F_H \amp F_D \amp F_V \amp F_{D'} \\ R_{1/4} \amp F_D \amp F_{D'} \amp F_H \amp F_V \amp R_{1/2} \amp R_{3/4} \amp R_0 \amp R_{1/4} \\ R_{1/2} \amp F_H \amp F_V \amp F_{D'} \amp F_D \amp R_{3/4} \amp R_0 \amp R_{1/4} \amp R_{1/2} \\ R_{3/4} \amp F_{D'} \amp F_D \amp F_V \amp F_H \amp R_0 \amp R_{1/4} \amp R_{1/2} \amp R_{3/4} \\ R_0 \amp F_V \amp F_H \amp F_D \amp F_{D'} \amp R_{1/4} \amp R_{1/2} \amp R_{3/4} \amp R_0 \end{array} \end{equation*}


text: Let’s look at how we can adapt the right mental model of useEffect to fix the exhaustive-deps warning.…


text:
The Index of EBSD data edit page

In previous chapters we have discussed how to select EBSD data by properties. In this chapter we are intested in the order the EBSD are stored within MTEX. Lets start by importing some sample data

and restricting it to very small rectangular subset

In the above plot each square corresponds to one entry in the variable ebsd. Lets visualize the order

We may easily select specific measurement pixels by specifying their indeces

Whether lines or columns run first is not related to MTEX but inherits from the ordering of the imported EBSD data. Since, we have restricted our large EBSD map to the small subset the indece of restricted data does not coincide with the indece of the imported data anymore. However, the original indeces are still stored in ebsd.id. Lets visualize them

In order to select EBSD data according to their original id use the option 'id', i.e.,

## Square Grids

In the cases of gridded data it is often useful to convert them into a matrix form.

This allows to select EBSD data simply by their coordinates within the grid, e.g., by

Note that the gridify command changes the order of measurements. They are now sort such that rows runs first and columns second, as this is the default convention how Matlab indexes matrices.

## Hexagonal Grid

The command gridify may also be applied to EBSD data measured on a hexagonal grid.

This rearranges the measurements in a matrix form which can be indexed similarly as in the square case.

Lets visualize the matrix coordinates for the hexagonal grid

## Cube Coordinates

In hexognal grids it is sometimes advantageous to use three digit cube coordinates to index the cell. This can be done using the commands hex2cube and cube2hex. Much more details on indexing hex grids can be found at here.


text:
# A Bit of Functional Programming in Rust, or A Misguided First Look at Rust for ML Programmers

Lately I’ve been playing more with rust, Mozilla’s new language for systems programming. Although it is a systems language, it is much more fun to use than languages like C and C++. My first impression was that it almost belonged in the ML family, but with slightly more complex memory management. However, as program complexity grows we see that Rust code takes a character of its own. Rust is built with safety, performance and concurrency in mind. Although the language is still not in a stable, backwards-compatible state (not until the 1.0 release), I feel it’s getting nearer to a state where your code does not break so often due to language changes. You can check the official tutorial for a good (although a bit idiosyncratic) introduction.

In this post I’ll show the results of attempting to write “pure” ML-style functional programming in Rust, introducing language features as needed so that people with previous functional programming experience can understand the code. Although the task will be easy in the beginning, showing a promising similarity to ML languages, as we try to go further it will become clear that Rust has its own character. This is an interesting perspective into Rust for a functional programmer like myself, I hope it’s useful to others as well.

# Functional Building Blocks

So what are the basic features of most functional languages? Recursion and lists. The old-school functional programming that evolved from “pure Lisp” can’t go anywhere without recursion and lists. Many books and courses about functional programming explain how to define and manipulate lists using recursion.

That’s what we’re doing here now, ML-style. In ML you’d define a list by creating an algebraic datatype with two variants: the empty list and a cons of an item and another list. To avoid introducing type variables (or, as the heathens call them, generics) just now, we’ll define a list that can contain only ints:

Although this should be familiar to ML or Haskell programmers, here we hit our first snag: what’s @IntList? Rust has manual memory management with optional garbage collection, so it distinguishes objects and pointers to objects. In the fully garbage-collected languages that are more common today, this difference has mostly disappeared under the surface (except for things like Java’s primitive types). In the IntList example, if we try to declare the Cons case with (int, IntList), the compiler will say to us:
 error: illegal recursive enum type; wrap the inner value in a box to make it representable
This means that IntList is trying to include a copy of itself inside it, and this can’t work because the size of an IntList is not known in advance. A “naked” IntList in rust is a value, not a reference. So to have a node in the list reference another, we have to make it a pointer. @IntList is a managed pointer to IntList. The object referenced by a managed pointer will be boxed, and its lifetime will be managed by the garbage collector.

Now we can define a function to find the length of an IntList:

Rust has ADTs, so it’s natural to also include pattern matching, as above. This is pretty standard ML-style functional code. A function to sum all the elements of a list would also be a direct translation from ML:

We can now test these functions:

The complete code, when compiled and run, will print the appropriate values (3 and 15), and we even have a small unit test to check that for us (rust has built-in support for tests). I know, the code to build the list isn’t very nice. We’ll solve that in a moment, but first let’s get rid of the monomorphic types.

## Polymorphism

The polymorphic (or generic if you prefer) list would be defined as

Naturally, List<A> is a list of elements of type A. Unfortunately, Rust uses C++’s syntax for type parameters (instead of, for example, square brackets like in Scala). Oh well, life’s not fair. Here is the polymorphic length function and a test:

By the way, the rust compiler translates polymorphism by monomorphisation, i.e. by generating specialized code for each instance of the type variable that is used by the program. In the above case it will generate code for function length over a list of strings. This is only superficially similar to C++ templates, in case you’re wondering; it’s rather like the process done by the SML compiler MLton.

## Functional Language?

At this point I imagine functional programmers are thinking two things: can we make length tail-recursive, and can we get sum_list back by defining a fold as a higher-order function? It’s easy to change length to be tail-recursive using the well-known accumulator pattern:

But this won’t do any good here: rust does not do tail call optimization. It’s unfortunate, but there are good reasons for that (summary: TCO clashes with some of Rust’s goals, including performance and C compatibility).

As for higher-order functions like fold, we could define them, but they would require language features that were not introduced yet (like the different kinds of pointers, or traits). There are closures in rust, but they’re a bit more complicated by virtue of the memory management. The thing is, although we can write this kind of functional-style code, it is not very idiomatic in Rust, and there are other ways to achieve the same goals.

## Macros

But there is still a promise to fulfill: a better way to build lists. It’s true: rust has macros. We can define a list building macro:

And then we can use the following code to build a List<int>:

But for this to work, we have to enable the macro_rules macro, as it’s considered an experimental feature. This can be done including #[feature(macro_rules)]; in the file. Here’s a complete example showing the use of the macro.

Rust’s macros are defined by example, and seem to be more limited than macro systems on other languages. For one, it’s always clear that a macro is being called instead of a function; also, the arguments to the macro must be inside parentheses. I’m not going to go into detail about the macro, but there’s a tutorial about them in the official documentation.

## In Conclusion

So after all this work defining lists using the language itself, many functional programmers would expect to see the “real”, built-in lists in the language. But there aren’t built-in functional-style lists in Rust, and as we saw here they don’t make for good, idiomatic Rust code.

We thus see that Rust isn’t ML. However, the influence from ML is clear, and the trait system comes from Haskell’s type classes. The specific design goals for Rust have made it diverge from standard statically-typed functional programming languages and go in a different direction.

Rust occupies a sparsely-populated area in the language design space, especially if you consider only the more mainstream languages (the memory management system owes a lot to Cyclone, a language very few people know). A Rust programmer gets more control, safety, and (possibly) better performance in exchange for having to think more about memory organisation and ownership. Is it worth it? For many applications maybe not, but there are domains where this kind of control is important. Also, as the language matures and people discover the best ways to write Rust code, it is expected that part of the complexity will be less exposed to most users. The exploration of this part of the design space will probably reveal better solutions than what is in place now. This is the natural evolution of the language, which is still in the beginning.

And in future posts I shall try to show good Rust code, or at least better code than what’s up here.

## Postscriptum

As I was writing this post, changes in the compiler made the managed pointers (the @ pointers) a gated feature, which means they should be explicitly switched on in the user code (I’ve changed the complete code examples in the gists accordingly). There have been talks of getting rid of special syntax for managed pointers for some time now, as a way to simplify the language. The original point stands, though: this is not idiomatic Rust code, and there are better ways to do the functional-style data manipulations we tried to do here.

• Mitch Skinner

Rust doesn’t have fully general tail-call optimization, but it does have sibling call optimization (tail calls with the same call signature):

So I think your length example would actually be TCOd.

• Andrei Formiga

Yes, but it’s different from TCO, so I decided to just leave it at that. This specific case may be optimized, but there are no guarantees. It’s good to note that some cases may be optimized, though.

• Jake

Minor correction

I think:
 enum List { Empty, Cons(int, @List) }

should be:

 enum List { Empty, Cons(A, @List) }

• Andrei Formiga

Indeed, it’s fixed now. Thanks for pointing it out.


text:
# Talk:Quasiperfect number

WikiProject Mathematics (Rated Stub-class, Low-priority)
This article is within the scope of WikiProject Mathematics, a collaborative effort to improve the coverage of Mathematics on Wikipedia. If you would like to participate, please visit the project page, where you can join the discussion and see a list of open tasks.
Mathematics rating:
 Stub Class
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Field: Number theory

I see no reason, in principle, why a qusiperfect number cannot be an even square or twice a square, which also have odd σ(n)'s. I would therefore include this in the request for citation. Septentrionalis 18:48, 2 October 2006 (UTC)

I can give a reason: Assume a quasiperfect number n σ(n) = 2n + 1 Now, n = 2^e k for some e and odd k, and ${\displaystyle \sigma (2^{e})=\,\!2^{e+1}-1}$

${\displaystyle (2^{e+1}\,\!-1)\sigma (k)=2^{e+1}k+1}$

Then, σ(k) is even unless k = l^2 for some l

${\displaystyle (2^{e+1}-1)\sigma (l^{2})=2^{e+1}l^{2}+\,\!1}$

Since σ(l^2) is an integer, we can write:

${\displaystyle 2^{e+1}l^{2}+1\,\!\equiv 0\ ({\mbox{mod}}\ 2^{e+1}-1)}$

Since

${\displaystyle 2^{e+1}\equiv 1\ ({\mbox{mod}}\ 2^{e+1}-1)}$

we can subtract 1 from both sides and arrive at

${\displaystyle l^{2}\equiv -1\ ({\mbox{mod}}\ 2^{e+1}-1)}$

Since l is an integer, we conclude that -1 is a quadratic residue of 2^(e+1) - 1. A familiar theorem from elementary number theory states that -1 can only be a quadratic residue of an integer r if r is of the form 4p + 1. Therefore, 2^(e+1) - 1 must be of the form 4p + 1 for some p; however, this is only true if e+1 = 1 and therefore e = 0, and n is an odd perfect square.

I can't remember exactly where I found that. 69.163.197.224 01:44, 9 November 2006 (UTC)

I'd wondered whether any such numbers exist. I was considering four cases of positive integers that equal the sum of their divisors:
• all divisors (the only such number is 1)
• all except 1 (which turn out to be the primes)
• all except itself (perfect numbers)
• all except 1 and itself
After I'd thought a bit about it and written a program to see if it finds any, I came here and found this name for them. By this point I had a simple proof that any odd quasiperfect must be square, but hadn't got anywhere near discovering that they can't be even. I wonder if there's a simpler proof of the latter. Meanwhile, I guess I'll have to have a look at that quadratic residue theorem. — Smjg (talk) 00:08, 15 September 2011 (UTC)

Refer to the above. If ${\displaystyle (2^{e+1}\,\!-1)\sigma (k)=2^{e+1}k+1}$ then ${\displaystyle \sigma (k)}$ is odd, not even. SophieAthena (talk) 22:47, 5 January 2013 (UTC)


text:
# 9.80 Table¶

The Table constructor provides a general structure for associative storage. This type provides hash tables in which data objects can be saved according to keys of any type. For a given table, specific types must be chosen for the keys and entries.

In this example the keys to the table are polynomials with integer coefficients. The entries in the table are strings.

t: Table(Polynomial Integer, String) := table()

 table()

Type: Table(Polynomial Integer,String)

To save an entry in the table, the seteltseteltTable operation is used. This can be called directly, giving the table a key and an entry.

setelt(t, x^2 - 1, "Easy to factor")

 “Easytofactor”

Type: String

Alternatively, you can use assignment syntax.

t(x^3 + 1) := "Harder to factor"

 “Hardertofactor”

Type: String

t(x) := "The easiest to factor"

 “Theeasiesttofactor”

Type: String

elt(t, x)

 “Theeasiesttofactor”

Type: String

This operation is called when a table is applied to a key using this or the following syntax.

t.x

 “Theeasiesttofactor”

Type: String

t x

 “Theeasiesttofactor”

Type: String

Parentheses are used only for grouping. They are needed if the key is an infixed expression.

t.(x^2 - 1)

 “Easytofactor”

Type: String

Note that the elteltTable operation is used only when the key is known to be in the table—otherwise an error is generated.

t (x^3 + 1)

 “Hardertofactor”

Type: String

You can get a list of all the keys to a table using the keyskeysTable operation.

keys t

 [x,x3+1,x2-1]

Type: List Polynomial Integer

If you wish to test whether a key is in a table, the searchsearchTable operation is used. This operation returns either an entry or “failed”.

search(x, t)

 “Theeasiesttofactor”

Type: Union(String,...)

search(x^2, t)

 “failed”

Type: Union(“failed”,...)

The return type is a union so the success of the search can be tested using case.

search(x^2, t) case "failed"

 true

Type: Boolean

The removeremoveTable operation is used to delete values from a table.

remove!(x^2-1, t)

 “Easytofactor”

Type: Union(String,...)

If an entry exists under the key, then it is returned. Otherwise removeremoveTable returns “failed”.

remove!(x-1, t)

 “failed”

Type: Union(“failed”,...)

The number of key-entry pairs can be found using the # #Table operation.

#t

 2

Type: PositiveInteger

Just as keyskeysTable returns a list of keys to the table, a list of all the entries can be obtained using the membersmembersTable operation.

members t

 [“Theeasiesttofactor”,”Hardertofactor”]

Type: List String

A number of useful operations take functions and map them on to the table to compute the result. Here we count the entries which have Hard as a prefix.

count(s: String +-> prefix?("Hard", s), t)

 1

Type: PositiveInteger

Other table types are provided to support various needs. \indent

AssociationList gives a list with a table view. This allows new entries to be appended onto the front of the list to cover up old entries. This is useful when table entries need to be stacked or when frequent list traversals are required. See AssociationListXmpPage for more information.

EqTable gives tables in which keys are considered equal only when they are in fact the same instance of a structure. See EqTableXmpPage for more information.

StringTable should be used when the keys are known to be strings. See StringTableXmpPage for more information.

SparseTable provides tables with default entries, so lookup never fails. The GeneralSparseTable constructor can be used to make any table type behave this way. See SparseTableXmpPage for more information.

KeyedAccessFile allows values to be saved in a file, accessed as a table. See KeyedAccessFileXmpPage for more information.


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Home > Sensing-Based Dynamic Spectrum Access Networks

# Sensing-Based Dynamic Spectrum Access Networks

Laneman awarded $473,132 by NSF to focus on a comprehensive approach to dynamic spectrum access based upon spectrum sensing. Investigator(s): J. Nicholas Laneman Sponsor: National Science Foundation Funding:$473,132

Timeframe:  9/11 to 8/14

Abstract:  The FCC released its National Broadband Plan citing the exponentially growing demand for mobile data services and the critical need to better utilize the radio spectrum.  Beyond retargeting certain frequency bands, the FCC is considering a paradigm shift of dynamic spectrum access (DSA) technology and policy.  DSA allows 'secondary' radios to transmit in underutilized 'white space' provided they create minimal interference to 'primary' radios.  Significant white space, as much as 85%, has been observed across time, frequency, and location.  However, even in the TV bands, the challenge of inconspicuous utilization of white spaces has not yet been achieved by the two leading solutions, distributed spectrum sensing and centralized emitter databases.

This project focuses on a comprehensive approach to DSA based upon spectrum sensing that combines the following in a feedback loop: novel modeling of primary and secondary transmissions; design of optimal spectrum sensing algorithms and secondary access protocols based upon these models; and experimental validation within a testbed of software-defined radios.  Preliminary results develop primary Markov models and optimal sequence detection algorithms for spectrum sensing that build upon the well-known Viterbi and forward-backward algorithms and expose fundamental limitations due to primary mismatch for the commonly used energy detector.  The connections to trellis- and graph-based algorithms from the channel coding community should introduce a sizable new toolbox to DSA researchers.  Secondary access protocols that take advantage of Markov process models for the primary users similarly exhibit improved performance tradeoffs between primary interference and secondary throughput.  Collaborators in industry and regulatory bodies will be kept informed of the research results with an aim toward impacting DSA technology and policy development at national and international levels.


text:
# Parabola separation queries and their application to stone throwing

2 VEGAS - Effective Geometric Algorithms for Surfaces and Visibility
INRIA Lorraine, LORIA - Laboratoire Lorrain de Recherche en Informatique et ses Applications
Abstract : Given two sets $A$ and $B$ of $m$ non-crossing line segments in the plane, we show how to compute in $O(m \log m)$ time a data structure that uses $O(m)$ storage and supports the following query in $O(\log m)$ time: Given a parabola $\p: y = ax^{2} + bx + c$, does $\p$ separate $A$ and $B$? This structure can be used to build a data structure that stores a simple polygon and allows ray-shooting queries along parabolic trajectories with vertical main axis. For a polygon of complexity~$n$, we can answer such stone-throwing'' queries in $O(\log^{2} n)$ time, using $O(n \log n)$ storage and $O(n \log^{2} n)$ preprocessing time. This matches the best known bound for circular ray shooting in simple polygons.
keyword :
Type de document :
Article dans une revue
International Journal of Computational Geometry and Applications, World Scientific Publishing, 2007, 17 (4), pp.349-360. 〈10.1142/S0218195907002379〉

Littérature citée [11 références]

https://hal.inria.fr/inria-00434090
Contributeur : Sylvain Lazard <>
Soumis le : vendredi 20 novembre 2009 - 16:57:00
Dernière modification le : jeudi 11 janvier 2018 - 06:20:14
Document(s) archivé(s) le : jeudi 17 juin 2010 - 19:05:15

### Fichier

ijcga-final.pdf
Fichiers produits par l'(les) auteur(s)

### Citation

Otfried Cheong, Hazel Everett, Hyo-Sil Kim, Sylvain Lazard, René Schott. Parabola separation queries and their application to stone throwing. International Journal of Computational Geometry and Applications, World Scientific Publishing, 2007, 17 (4), pp.349-360. 〈10.1142/S0218195907002379〉. 〈inria-00434090〉

### Métriques

Consultations de la notice

## 396

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text:
# SectionILTInjective Linear Transformations¶ permalink

Some linear transformations possess one, or both, of two key properties, which go by the names injective and surjective. We will see that they are closely related to ideas like linear independence and spanning, and subspaces like the null space and the column space. In this section we will define an injective linear transformation and analyze the resulting consequences. The next section will do the same for the surjective property. In the final section of this chapter we will see what happens when we have the two properties simultaneously.

# SubsectionILTInjective Linear Transformations¶ permalink

As usual, we lead with a definition.

##### DefinitionILTInjective Linear Transformation

Suppose $\ltdefn{T}{U}{V}$ is a linear transformation. Then $T$ is injective if whenever $\lteval{T}{\vect{x}}=\lteval{T}{\vect{y}}\text{,}$ then $\vect{x}=\vect{y}\text{.}$

Given an arbitrary function, it is possible for two different inputs to yield the same output (think about the function $f(x)=x^2$ and the inputs $x=3$ and $x=-3$). For an injective function, this never happens. If we have equal outputs ($\lteval{T}{\vect{x}}=\lteval{T}{\vect{y}}$) then we must have achieved those equal outputs by employing equal inputs ($\vect{x}=\vect{y}$). Some authors prefer the term one-to-one where we use injective, and we will sometimes refer to an injective linear transformation as an injection.

# SubsectionEILTExamples of Injective Linear Transformations¶ permalink

It is perhaps most instructive to examine a linear transformation that is not injective first.

Here is a cartoon of a non-injective linear transformation. Notice that the central feature of this cartoon is that $\lteval{T}{\vect{u}}=\vect{v}=\lteval{T}{\vect{w}}\text{.}$ Even though this happens again with some unnamed vectors, it only takes one occurrence to destroy the possibility of injectivity. Note also that the two vectors displayed in the bottom of $V$ have no bearing, either way, on the injectivity of $T\text{.}$

To show that a linear transformation is not injective, it is enough to find a single pair of inputs that get sent to the identical output, as in Example NIAQ. However, to show that a linear transformation is injective we must establish that this coincidence of outputs never occurs. Here is an example that shows how to establish this.

Here is the cartoon for an injective linear transformation. It is meant to suggest that we never have two inputs associated with a single output. Again, the two lonely vectors at the bottom of $V$ have no bearing either way on the injectivity of $T\text{.}$

Let us now examine an injective linear transformation between abstract vector spaces.

# SubsectionKLTKernel of a Linear Transformation¶ permalink

For a linear transformation $\ltdefn{T}{U}{V}\text{,}$ the kernel is a subset of the domain $U\text{.}$ Informally, it is the set of all inputs that the transformation sends to the zero vector of the codomain. It will have some natural connections with the null space of a matrix, so we will keep the same notation, and if you think about your objects, then there should be little confusion. Here is the careful definition.

##### DefinitionKLTKernel of a Linear Transformation

Suppose $\ltdefn{T}{U}{V}$ is a linear transformation. Then the kernel of $T$ is the set \begin{equation*} \krn{T}=\setparts{\vect{u}\in U}{\lteval{T}{\vect{u}}=\zerovector}\text{.} \end{equation*}

Notice that the kernel of $T$ is just the preimage of $\zerovector\text{,}$ $\preimage{T}{\zerovector}$ (Definition PI). Here is an example.

We know that the span of a set of vectors is always a subspace (Theorem SSS), so the kernel computed in Example NKAO is also a subspace. This is no accident, the kernel of a linear transformation is always a subspace.

##### Proof

Let us compute another kernel, now that we know in advance that it will be a subspace.

Our next theorem says that if a preimage is a nonempty set then we can construct it by picking any one element and adding on elements of the kernel.

##### Proof

This theorem, and its proof, should remind you very much of Theorem PSPHS. Additionally, you might go back and review Example SPIAS. Can you tell now which is the only preimage to be a subspace?

Here is the cartoon which describes the “many-to-one” behavior of a typical linear transformation. Presume that $\lteval{T}{\vect{u}_i}=\vect{v}_i\text{,}$ for $i=1,2,3\text{,}$ and as guaranteed by Theorem LTTZZ, $\lteval{T}{\zerovector_U}=\zerovector_V\text{.}$ Then four pre-images are depicted, each labeled slightly different. $\preimage{T}{\vect{v}_2}$ is the most general, employing Theorem KPI to provide two equal descriptions of the set. The most unusual is $\preimage{T}{\zerovector_V}$ which is equal to the kernel, $\krn{T}\text{,}$ and hence is a subspace (by Theorem KLTS). The subdivisions of the domain, $U\text{,}$ are meant to suggest the partioning of the domain by the collection of pre-images. It also suggests that each pre-image is of similar size or structure, since each is a “shifted” copy of the kernel. Notice that we cannot speak of the dimension of a pre-image, since it is almost never a subspace. Also notice that $\vect{x},\,\vect{y}\in V$ are elements of the codomain with empty pre-images.

The next theorem is one we will cite frequently, as it characterizes injections by the size of the kernel.

##### Proof

You might begin to think about how Figure 7.48 would change if the linear transformation is injective, which would make the kernel trivial by Theorem KILT.

Click to open

# SubsectionILTLIInjective Linear Transformations and Linear Independence¶ permalink

There is a connection between injective linear transformations and linearly independent sets that we will make precise in the next two theorems. However, more informally, we can get a feel for this connection when we think about how each property is defined. A set of vectors is linearly independent if the only relation of linear dependence is the trivial one. A linear transformation is injective if the only way two input vectors can produce the same output is in the trivial way, when both input vectors are equal.

# SubsectionILTDInjective Linear Transformations and Dimension¶ permalink

##### Proof

Notice that the previous example made no use of the actual formula defining the function. Merely a comparison of the dimensions of the domain and codomain are enough to conclude that the linear transformation is not injective. Archetype M and Archetype N are two more examples of linear transformations that have “big” domains and “small” codomains, resulting in “collisions” of outputs and thus are non-injective linear transformations.

# SubsectionCILTComposition of Injective Linear Transformations¶ permalink

In Subsection LT.NLTFO we saw how to combine linear transformations to build new linear transformations, specifically, how to build the composition of two linear transformations (Definition LTC). It will be useful later to know that the composition of injective linear transformations is again injective, so we prove that here.

Click to open

##### 1

Suppose $\ltdefn{T}{\complex{8}}{\complex{5}}$ is a linear transformation. Why is $T$ not injective?

##### 2

Describe the kernel of an injective linear transformation.

# SubsectionExercises

##### C10

Each archetype below is a linear transformation. Compute the kernel for each.

Archetype M, Archetype N, Archetype O, Archetype P, Archetype Q, Archetype R, Archetype S, Archetype T, Archetype U, Archetype V, Archetype W, Archetype X

##### C20

The linear transformation $\ltdefn{T}{\complex{4}}{\complex{3}}$ is not injective. Find two inputs $\vect{x},\,\vect{y}\in\complex{4}$ that yield the same output (that is $\lteval{T}{\vect{x}}=\lteval{T}{\vect{y}}$). \begin{equation*} \lteval{T}{\colvector{x_1\\x_2\\x_3\\x_4}}= \colvector{ 2x_1+x_2+x_3\\ -x_1+3x_2+x_3-x_4\\ 3x_1+x_2+2x_3-2x_4 }\text{.} \end{equation*}

Solution
##### C25

Define the linear transformation \begin{equation*} \ltdefn{T}{\complex{3}}{\complex{2}},\quad \lteval{T}{\colvector{x_1\\x_2\\x_3}}=\colvector{2x_1-x_2+5x_3\\-4x_1+2x_2-10x_3}\text{.} \end{equation*} Find a basis for the kernel of $T\text{,}$ $\krn{T}\text{.}$ Is $T$ injective?

Solution
##### C26

Let \begin{equation*} A = \begin{bmatrix} 1 & 2 & 3 & 1 & 0\\ 2 & -1 & 1 & 0 & 1\\ 1 & 2 & -1 & -2 & 1\\ 1 & 3 & 2 & 1 & 2 \end{bmatrix} \end{equation*} and let $\ltdefn{T}{\complex{5}}{\complex{4}}$ be given by $\lteval{T}{\vect{x}}=A\vect{x}\text{.}$ Is $T$ injective? (Hint: No calculation is required.)

Solution
##### C27

Let $\ltdefn{T}{\complex{3}}{\complex{3}}$ be given by $\lteval{T}{\colvector{x\\y\\z}} = \colvector{2x + y + z\\ x - y + 2z\\ x + 2y - z}\text{.}$ Find $\krn{T}\text{.}$ Is $T$ injective?

Solution
##### C28

Let \begin{equation*} A = \begin{bmatrix} 1 & 2 & 3 & 1 \\ 2 & -1 & 1 & 0 \\ 1 & 2 & -1 & -2 \\ 1 & 3 & 2 & 1 \end{bmatrix} \end{equation*} and let $\ltdefn{T}{\complex{4}}{\complex{4}}$ be given by $\lteval{T}{\vect{x}}=A\vect{x}\text{.}$ Find $\krn{T}\text{.}$ Is $T$ injective?

Solution
##### C29

Let \begin{equation*} A = \begin{bmatrix} 1 & 2 & 1 & 1 \\ 2 & 1 & 1 & 0 \\ 1 & 2 & 1 & 2 \\ 1 & 2 & 1 & 1 \end{bmatrix} \end{equation*} and let $\ltdefn{T}{\complex{4}}{\complex{4}}$ be given by $\lteval{T}{\vect{x}}=A\vect{x}\text{.}$ Find $\krn{T}\text{.}$ Is $T$ injective?

Solution
##### C30

Let $T : M_{22} \rightarrow P_2$ be given by $T\left(\begin{bmatrix} a & b \\ c & d \end{bmatrix}\right) = (a + b) + (a + c)x + (a + d)x^2\text{.}$ Is $T$ injective? Find $\krn{T}\text{.}$

Solution
##### C31

Given that the linear transformation $\ltdefn{T}{\complex{3}}{\complex{3}}\text{,}$ $\lteval{T}{\colvector{x\\y\\z}} = \colvector{2x + y\\2y + z\\x + 2z}$ is injective, show directly that \begin{equation*} \set{ \lteval{T}{\vect{e}_1},\, \lteval{T}{\vect{e}_2},\, \lteval{T}{\vect{e}_3} } \end{equation*} is a linearly independent set.

Solution
##### C32

Given that the linear transformation $\ltdefn{T}{\complex{2}}{\complex{3}}\text{,}$ $\lteval{T}{\colvector{x\\y}} = \colvector{x+y\\2x + y\\x + 2y}$ is injective, show directly that \begin{equation*} \set{ \lteval{T}{\vect{e}_1},\, \lteval{T}{\vect{e}_2} } \end{equation*} is a linearly independent set.

Solution
##### C33

Given that the linear transformation $\ltdefn{T}{\complex{3}}{\complex{5}}\text{,}$ \begin{equation*} \lteval{T}{\colvector{x\\y\\z}} = \begin{bmatrix} 1 & 3 & 2\\ 0 & 1 & 1\\ 1 & 2 & 1\\ 1 & 0 & 1\\ 3 & 1 & 2 \end{bmatrix} \colvector{x\\y\\z} \end{equation*} is injective, show directly that \begin{equation*} \set{ \lteval{T}{\vect{e}_1},\, \lteval{T}{\vect{e}_2},\, \lteval{T}{\vect{e}_3} } \end{equation*} is a linearly independent set.

Solution
##### C40

Show that the linear transformation $R$ is not injective by finding two different elements of the domain, $\vect{x}$ and $\vect{y}\text{,}$ such that $\lteval{R}{\vect{x}}=\lteval{R}{\vect{y}}\text{.}$ ($S_{22}$ is the vector space of symmetric $2\times 2$ matrices.) \begin{equation*} \ltdefn{R}{S_{22}}{P_1}\quad \lteval{R}{\begin{bmatrix}a&b\\b&c\end{bmatrix}}=(2a-b+c)+(a+b+2c)x\text{.} \end{equation*}

Solution
##### M60

Suppose $U$ and $V$ are vector spaces. Define the function $\ltdefn{Z}{U}{V}$ by $\lteval{Z}{\vect{u}}=\zerovector_{V}$ for every $\vect{u}\in U\text{.}$ Then by Exercise LT.M60, $Z$ is a linear transformation. Formulate a condition on $U$ that is equivalent to $Z$ being an injective linear transformation. In other words, fill in the blank to complete the following statement (and then give a proof): $Z$ is injective if and only if $U$ is . (See Exercise SLT.M60, Exercise IVLT.M60.)

##### T10

Suppose $\ltdefn{T}{U}{V}$ is a linear transformation. For which vectors $\vect{v}\in V$ is $\preimage{T}{\vect{v}}$ a subspace of $U\text{?}$

Solution
##### T15

Suppose that that $\ltdefn{T}{U}{V}$ and $\ltdefn{S}{V}{W}$ are linear transformations. Prove the following relationship between kernels. \begin{equation*} \krn{T}\subseteq\krn{\compose{S}{T}}\text{.} \end{equation*}

Solution
##### T20

Suppose that $A$ is an $m\times n$ matrix. Define the linear transformation $T$ by \begin{equation*} \ltdefn{T}{\complex{n}}{\complex{m}},\quad \lteval{T}{\vect{x}}=A\vect{x}\text{.} \end{equation*} Prove that the kernel of $T$ equals the null space of $A\text{,}$ $\krn{T}=\nsp{A}\text{.}$

Solution


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# Is the Fundamental Theorem of Calculus really applicable to the definition of work?

When the force $F$ on an object is not constant, then the work it performs is defined as $$W = \int_{x_0}^{x} F(X)dX.$$

Now, the Fundamental Theorem of Calculus states that

$$\text{If}\,\,\, f(x) = \int_{x_0}^{x} F(u)du, \,\,\,\text{then}\,\,\, \dfrac{d{f(x)}}{dx} = F(x).$$

If it is true (definitely it is), then I can say that the rate of change of work with respect to displacement is force or that the instantaneous rate of change of work with respect to displacement is force? I find this ridiculous! Can work be instantaneous? I can't imagine how work can be instantaneous. If the statement is wrong, how can the Fundamental Theorem of Calculus be true? Please help.

The Fundamental Theorem of Calculus is of course correct, and you are applying it correctly. The statements

• the rate of change of work with respect to displacement is force

and

• the instantaneous rate of change of work with respect to displacement is force

are correct. The thing to keep in mind is that it's not work that is instantaneous, but its rate of change. The work performed on the system is constantly changing, and it is performed continuously through time (and thence over displacement). Its derivative with respect do displacement is simply how fast it is changing, and this is a function of the particular instant as much as the work itself, and also the displacement, velocity, and so on.

There is an additional consequence to this interpretation: the force in an object is the work you would need to perform on an object to push it a unit displacement in the given direction. This is correct, though it is indeed a little mind-bending! If it's any help, this will get trumped by things like the principle of virtual work.

• +1 . Does instantaneous velocity exist? What about Zeno's paradox?? Mr. Feynman replied,Calculus gives the answer. And what about intuition?? ...if the instantaneous velocity is $v$ km/hr , it really does mean that if from that point, the motion were uniform, the body would have velocity $v$ after 1 hr. – user36790 Nov 24 '14 at 11:54
• Sir, just an example is needed! Can you please give?? I will be grateful to you. – user36790 Nov 24 '14 at 14:38
• You can directly translate: "If the instantaneous velocity is $v$ m/s, it really does mean that, if from that moment the motion were uniform, the body would have advanced $v$ m in 1s." This goes directly into "If the instantaneous-rate-of-change-of-work (i.e. force) is $F$ N, it really does mean that, if from that moment the motion were uniform, the external force would have performed $F$ joules of work over the next 1m." – Emilio Pisanty Nov 24 '14 at 14:58
• +1 . Really a great answer and, of course, the comment. Sir, in the same way can I apply FOTC in case of moment of inertia?? $$dI = r^2 .dm$$ . What will be the instantaneous rate of change of $I$ with respect to mass?? Thanks for clearing my confusion:) – user36790 Nov 24 '14 at 16:59
• That one is a bit harder. The interpretation of that integral goes something as "adding a unit mass of $m$ kg at squared-radius $r^2$ m$^2$ will increase the moment of inertia by $m r^2$ kg m$^2$". But in general it doesn't serve much to go around looking for every possible integral you might apply the FTC to. – Emilio Pisanty Nov 24 '14 at 17:49

No matter how ridiculous you may find it, it is true. The most general definition of work is indeed that the infinitesimal work done along an infinitesimal path is just force times the length of the path, i.e.

$$\mathrm{d}W = F\mathrm{d}s$$

Therefore, the amount of work done along a path in space $\gamma : [a,b] \to \mathbb{R}^3$ is the line integral

$$W = \int_\gamma \vec F(\vec x) \cdot \mathrm{d}\vec x$$

which, if the force is a conservative force, i.e. if there exists a function $U : \mathbb{R}^3 \to \mathbb{R}$ such that $\vec F = \vec \nabla U$, reduces to

$$W = U(\gamma(b)) - U(\gamma(a))$$

so that, for conservative forces in one dimension, indeed $\frac{\mathrm{d}W}{\mathrm{d}x} = F$.


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ContextMenu[CurrentContext][Entries] - Maple Help

# Online Help

###### All Products    Maple    MapleSim

Home : Support : Online Help : System : Context-Sensitive Menus : CurrentContext : Entries : ContextMenu/CurrentContext/Entries/Remove

ContextMenu[CurrentContext][Entries]

 Remove
 Remove an entry in the context menu

 Calling Sequence Entries[Remove](action)

Parameters

 action - string or list(string); menu entry to be removed

Description

 • The Entries[Remove] command removes all references to an entry from the context menu for the current session.  The effect of this command is not permanent unless you save the menu to a repository using ContextMenu[Save].
 • The parameter action specifies the action or actions to remove. If action is a string, any entry in the top-level context menu with this name will be removed.  If action is a list of strings, this is interpreted as a hierarchical sequence of submenus starting from the top-level menu.  Any menu entries whose submenu entry sequence begins with action will be removed.

Examples of Entries[Remove]

 > with(ContextMenu[CurrentContext]):
 Remove the top-level menu entry that negates a boolean value.
 > Entries[Remove]("Negate");
 Remove the submenu entry that sorts a list in ascending order.
 > Entries[Remove](["Sorts", "Ascending Order"]);

 See Also

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# What is the average speed of an object that is still at t=0 and accelerates at a rate of a(t) =t^2-4 from t in [2, 3]?

Jan 12, 2018

4.58 $\frac{m}{\sec}$

#### Explanation:

Acceleration of the object is given as,$a = 4 {t}^{2} - 4$
So,velocity will be $\frac{{t}^{3}}{3} - 4 t$ (by integrating)
Hence,displacement will be,
$S = \frac{{t}^{4}}{12} - \left(2 {t}^{2}\right)$ (by integrating)
Hence,displacement with in t=2 to t=3 will be -4.58 meter
Now here distance=displacement that has happened in between 2 sec and 3 sec
Hence average speed with in this duration=(total distance covered/total time)
I.e $\left(\frac{4.58}{1}\right)$ $\frac{m}{\sec}$ or 4.58 $\frac{m}{\sec}$


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# Chapter 2 - Solving Equations - 2-4 Solving Equations with Variables on Both Sides - Practice and Problem-Solving Exercises: 29

No Solutions

#### Work Step by Step

$-2k = -2k + 5$ Add $2k$ to each side of the equation to get : $0 = 5$ $0 \ne 5$, so there are no solutions of k for which $-2k = -2k + 5$.

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