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User eric finster - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T08:39:16Z http://mathoverflow.net/feeds/user/4466 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/37992/the-homotopy-cofiber-of-the-smash-product-of-two-maps-of-spectra The homotopy cofiber of the smash product of two maps of spectra Eric Finster 2010-09-07T18:47:30Z 2011-11-24T02:29:43Z
It is a standard fact that smashing with a fixed spectrum $Z$ preserves cofiber sequences. So if I have a cofiber sequence $$X \xrightarrow{f} Y \rightarrow C_f$$ then there is also a cofiber sequence $$Z \wedge X \rightarrow Z \wedge Y \rightarrow Z \wedge C_f$$

If more generally I have a map $Z \xrightarrow{g} W$, is there any formula for the cofiber of the map $$Z \wedge X \xrightarrow{g \wedge f} W \wedge Y$$ in terms of $C_f$ and $C_g$? (The above discussion corresponding to $g = \mathrm{id}_Z$).
 http://mathoverflow.net/questions/46083/controlling-reflective-subcategories-and-localizations Controlling Reflective Subcategories and Localizations Eric Finster 2010-11-14T22:55:48Z 2010-11-15T00:58:38Z
Localizations are an extremely important part of modern homotopy theory. Both the category of spaces an spectra have a plethora of interesting localizations: at a fixed prime, rational, with respect to a given homology theory, etc.

I started wondering if it's possible to create "toy models" of some kind where the collection of possible localizations is somehow under control in a nice way. This would seem very difficult to accomplish using model categories, since a major theme of the model category literature is that there are almost always lots of localizations around. It's possible one could do something with $\infty$-categories, since the notion of localization makes sense there as well, but for now I'll just stick to ordinary categories for simplicity. In fact, localizations are just special reflective subcategories, so let's just think about those.

So suppose I have a category $\mathcal C$. Can I always find a new category which has $\mathcal C$ as a (proper, so that this is nontrivial) reflective subcategory? Yes, though I'm not very happy with my solution. Let $\mathcal C_0$ be the set of objects of $\mathcal C$ regarded as a discrete category. The inclusion $\mathcal C_0 \rightarrow \mathcal C$ is a diagram in $\mathcal Cat$, and it's easy to see that the Grothendieck construction on this diagram has $\mathcal C$ as a reflective subcategory, albeit in a rather trivial way. Are there other (possibly better) examples?

Now what if I have a collection of categories $\mathcal C_i$ for $i \in \mathcal I$. Is it possible to build a new category which contains each of the $\mathcal C_i$ as a reflective subcategory?

In general are there constructions which affect the collection of reflective subcategories in predictable ways? For example, it's easy to see that in the disjoint union $\mathcal C \coprod \mathcal D$ of two categories, the collection of reflective subcategories is just the product of those of $\mathcal C$ and those of $\mathcal D$. The cartesian product $\mathcal C \times \mathcal D$ seems a bit harder to understand.

I'd be interested to hear about homotopy theoretic variants as well. It just seemed easier to me to think about categories first.
 http://mathoverflow.net/questions/46079/do-quasi-categories-have-a-completion/46085#46085 Answer by Eric Finster for Do quasi-categories have a `completion'? Eric Finster 2010-11-14T23:10:14Z 2010-11-14T23:10:14Z
Maybe this is not exactly what you are looking for, but it might be a place to start.

There is an analog of the Yoneda embedding for quasi-categories which takes the form $$j : \mathcal C \rightarrow Map(\mathcal C^{op}, \mathcal S)$$ where here $\mathcal S$ is the quasicategory of spaces (that is, the coherent nerve of the category of Kan complexes.) This map is an embedding which can be regarded as the free completion of $\mathcal C$ with respect to colimits, just as in ordinary category theory, and moreover preserves all limits which exist in $\mathcal C$. This is of course quite a bit more than just finitely bicomplete, but Lurie's book on Higher Topos Theory outlines a method for adjoining any class of colimits, and presumably this would work for finite colimits as well if that's what you want.
 http://mathoverflow.net/questions/41264/equivalence-of-cosimplicial-models-for-homotopy-pullbacks Equivalence of cosimplicial models for homotopy pullbacks Eric Finster 2010-10-06T12:41:08Z 2010-10-06T19:09:01Z
Given a diagram $X_1 \rightarrow X_{12} \leftarrow X_2$ of spaces (though I think the question applies more generally), there are two cosimplicial resolutions which I've seen used to compute the homotopy pullback. The first one, which I'll call $\mathcal A$ is given by $$\mathcal A^p = X_1 \times X_{12}^{\times p} \times X_2$$ and the coface maps are like in a bar (cobar?) construction. This is the one you see, for example, in the construction of the Eilenberg-Moore spectral sequence.

For the other one, let's let $\mathscr P$ denote the pullback category, and $F$ our functor. Then the standard cosimplicial replacement, which I'll write $\mathcal B$, is given by $$\mathcal B^p = \prod_{\substack{x_0 \rightarrow \cdots \rightarrow x_p \ \in \mathscr P}} F(x_p)$$

It seems that I always encounter one or the other, but not both, which leads me to my question: how do I know they give the same answer? The first one computes the homotopy pullback basically by definition, whereas the second one is a bit mysterious to me. Is there a simple way to compare these two cosimplicial spaces? They do not appear to be level-wise equivalent. Can we say anything about the relationship of their $\mathrm{Tot}$ towers?
 http://mathoverflow.net/questions/37647/the-dold-thom-theorem-for-infinity-categories The Dold-Thom theorem for infinity categories? Eric Finster 2010-09-03T18:40:59Z 2010-09-06T15:20:50Z
Let $\mathcal{M}$ denote the category of finite sets and monomorphisms, and let $\mathcal T$ denote the category of based spaces. For a based space $X \in \mathcal T$, one has a canonical funtor $S_X : \mathcal M \rightarrow \mathcal T$ defined by $\{n\} \mapsto X^n$. The definition on morphisms is to insert basepoints on the factors which are not in the image of a given monomorphism.

As is well know, the homotopy groups of $\mathrm{colim} S_X = SP^\infty X$ give the homology of $X$ (this is the Dold-Thom theorem), and the homotopy groups of $\mathrm{hocolim} S_X = SP^\infty_h X$ given the stable homotopy of $X$.

Is there a model for $SP^\infty X$, the ordinary infinite symmetric product, as a homotopy colimit as opposed to a categorical colimit?

The motivation for this question comes from thinking about $\infty$-categories. In an $\infty$-category, one does not really have a good notion (at least not one that I am aware of) of strict categorical colimits. So I'm wondering if there is, nonetheless, some easily defined functor on the $\infty$-category of spaces which will let us calculate ordinary homology. In short, is there any $\infty$-categorical analog of the Dold-Thom theorem?

Update: Following up on André's remark it seems using the orbit category is heading in the right direction, at least for the $n$-th approximations. I'll just quickly sketch what I have so far:

Let $\mathcal O(\Sigma_n)$ denote the orbit category. The objects are the homogeneous (discrete) spaces $\Sigma_n/H$ (with left actions) as $H$ runs over all the subgroups of $\Sigma_n$, and the morphisms are the $\Sigma_n$-equivariant maps. There is a canonical functor $$\Sigma_n \rightarrow \mathcal O(\Sigma_n)^{op}$$ where we regard $\Sigma_n$ as a category with one object as usual.

Given a $\Sigma_n$ space $X$, right Kan extension along this inclusion produces a $\mathcal O(\Sigma_n)^{op}$ diagram $\tilde X$ defined by $$\tilde X(\Sigma_n/H) = X^H$$ It turns out that the above inclusion is final so that it induces an isomorphism of colimits. Hence $\mathrm{colim}_{\mathcal O(\Sigma_n)} \tilde X \cong X_{\Sigma_n}$, i.e., the coinvariants. It's also not hard to see that the undercategories are copies of $B\Sigma_n$, hence not contractible, so we don't expect an equivalence of homotopy colimits, which is good.

On the other hand, I can now show that when $X$ is discrete, the canonical map $$\mathrm{hocolim} \tilde X \rightarrow \mathrm{colim} \tilde X$$ is an equivalence. My methods here do not generalize to all spaces, so if someone has a reference for why this is true in general, that would be much appreciated. (I think something like this must appear in May's book on equivariant homotopy theory if it's true, but I did not have it available this weekend.)

The remaining part would be to let $n \rightarrow \infty$, but somehow this seems like it should not be too bad. (Something like: make a functor $\mathcal M \rightarrow \mathcal Cat$ by $n \mapsto \mathcal O(\Sigma_n)$. Take the Grothendieck construction. Some natural diagram on this category might give the right answer.)
 http://mathoverflow.net/questions/37647/the-dold-thom-theorem-for-infinity-categories/37898#37898 Answer by Eric Finster for The Dold-Thom theorem for infinity categories? Eric Finster 2010-09-06T15:20:50Z 2010-09-06T15:20:50Z
It so happens that Emmanuel Dror Farjoun is visiting the EPFL this week. I figured I'd ask him about this problem at lunch today. What a coincidence! He proved exactly this statement using the exact same techniques. In fact, the construction of $SP^n$ as a homotopy colimit is the subject of Chapter 4 in "Cellular Spaces, Null Spaces, and Homotopy Localization," Lecture Notes in Mathematics, 1622.

It turns out, the idea works more generally so that we can always replace strict colimits with homotopy colimits: define an orbit on a category $\mathcal C$ to be a functor $O : \mathcal C \rightarrow \mathcal Set$ such that $\mathrm{colim}_{\mathcal C} O \cong *$. There is a category of such functors which we call the orbit category of $\mathcal C$, denoted $\mathcal O(\mathcal C)$. The Yoneda embedding factors through $\mathcal O(\mathcal C)$, and the right Kan extension along this inclusion always results in a "free" diagram.

I still want to play around with the construction a bit to see if there are any wrinkles with $n \rightarrow \infty$, and if I can use this to give easy calculations of homology in the $\infty$-category $\mathcal S$, but I think it's safe to say at this point that answer to my question is yes.
 http://mathoverflow.net/questions/36230/does-f-have-a-description-in-terms-of-the-grothendieck-construction/36317#36317 Answer by Eric Finster for Does F_* have a description in terms of the Grothendieck construction? Eric Finster 2010-08-21T19:20:11Z 2010-08-21T19:20:11Z
I guess it depends what you would accept as a "description in terms of the Grothendieck construction."

For each $d \in D$, we have the projection $\pi_d : d/F \rightarrow C$. Given some $\gamma : C \rightarrow Set$, I can form the composite $$\gamma \circ \pi_d : d/F \rightarrow C \rightarrow Set$$

This determines a discrete opfibration $\int \gamma \circ \pi_d$ over $d/F$. Let $\Gamma_{d/F}$ denote the set of sections of the projection $\int \gamma \circ \pi_d \rightarrow d/F$. Then I believe you will find that $F_* \gamma$ is the Grothendieck construction on the functor $d \mapsto \Gamma_{d/F}$.

Of course, this is not really that interesting, since it is really nothing more than the observation that the limit of a functor $F : C \rightarrow Set$ can be calculated as the set of sections of the natural projection $\int F \rightarrow C$, together with definition of the right Kan extension.

On the other hand, it suggests (to me at least) that there is unlikely to be a more "global" description in terms of the Grothendieck construction since the object we are trying to describe is like a "union of a collection of maps," and these to operations tend not to commute (maps from a union is the product of the maps on each component). Probably you knew all this, but maybe somebody will find it useful . . .
 http://mathoverflow.net/questions/36048/modules-and-square-zero-extensions Modules and Square Zero Extensions Eric Finster 2010-08-19T02:23:57Z 2010-08-19T03:27:14Z
Let $R$ be a commutative ring, $RMod$ its category of modules and $CRing$ the category of commutative rings.

There's an embedding $RMod \rightarrow CRing/R$ that sends an $R$-module $M$ to the ring $$R \oplus M$$ (the direct sum taken as modules) with multiplication $(r_0,m_0)(r_1,m_1) = (r_0 r_1, r_0 m_1 + r_1 m_0)$. This functor restricts to an equivalence of categories between $RMod$ and $Ab(CRing/R)$, the category of abelian group objects in the slice category.

The projection $R \oplus M \rightarrow R$ makes this ring into a square-zero extension of $R$. My understanding is that in algebraic geometry, one thinks of a square zero extension of a ring as a kind of infinitesimal extension of $Spec (R)$. So the category of $R$-modules can be viewed geometrically as parameterizing a certain class of infinitesimal objects related to $R$.

On the other hand, of course, the category $RMod$ is equivalent to the category of quasicoherent sheaves on $Spec(R)$, which seems, to me at least, totally unrelated to my previous description.

So my question is: are these two views of the same category somehow related? When I think of the sheaf associated to a module $M$, does it somehow contain information about the corresponding infinitesimal extension? What about when I look at cohomology with coefficients in that sheaf?
 http://mathoverflow.net/questions/36016/understanding-the-etale-space-construction-from-a-formal-viewpoint/36028#36028 Answer by Eric Finster for Understanding the etale space construction from a formal viewpoint Eric Finster 2010-08-18T21:49:18Z 2010-08-19T00:28:48Z
Here is a sketch of why I think the condition that $Y$ is a local homeomorphism over $X$ should be sufficient for the counit to be a homeomorphism. I haven't worked out the converse yet.

For a presheaf $F \in Set^{\mathcal{O}(X)^{op}}$, the formula for the left Kan extension should be $$L(F) = \mathrm{colim}_{y(U) \rightarrow F} U$$ where $y : \mathcal{O}(X) \rightarrow Set^{\mathcal{O}(X)^{op}}$ is the Yoneda embedding. By Yoneda's lemma, the indexing category for the colimit is exactly the category of elements of $F$ which I will write as $\int F$.

Now, consider the case where $F = \Gamma_Y$ for some space $p: Y \rightarrow X$ and assume that $p$ is a local homeomorphism. We have $\Gamma_Y(U) = \{\sigma : U \rightarrow Y | p \circ \sigma = \mathrm{id}_U \}$. So the objects in the category $\int \Gamma_Y$ are exactly the sections over the various open sets of $X$, and the morphisms are given by restriction of sections. I'll write $d(\sigma)$ for the domain of a given section.

Our Kan extension formula becomes $$L(\Gamma_Y) = \mathrm{colim}_{\sigma \in \int \Gamma_Y} d(\sigma)$$ From here it's easy to see what the counit is: since our object is given by a colimit, it suffices to construct a map $d(\sigma) \rightarrow Y$ for each $\sigma \in \int \Gamma_Y$. But clearly $\sigma$ itself qualifies.

Now choose an open covering $\{V_\alpha\}$ of $Y$ such that $p$ restricts to a homeomorphism on each $V_\alpha$. We then have a collection $\{\sigma_\alpha : p(V_\alpha) \rightarrow V_\alpha\}$ of sections by choosing the inverse to each restriction. My claim would be that this collection is cofinal (or final? I can never remember which) in the category $\int \Gamma_Y$ so that we can restrict our colimit to just this subcategory. Notice that in this case, the components of the counit map above are homeomorphisms.

Moreover, this subcategory should also be cofinal in $\mathcal{O}(Y)$ by associating $\sigma_\alpha$ with the open set $V_\alpha$. Then the fact that the counit is a homeomorphism should be the statement that a topological space is the colimit of any of its open coverings.

Is this along the lines of what you were thinking?
 http://mathoverflow.net/questions/34951/a-transformation-of-infinite-series A transformation of infinite series Eric Finster 2010-08-08T21:35:20Z 2010-08-09T09:09:42Z
Suppose I have a convergent infinite series $\sum_{n=0}^\infty (-1)^n a_n = S_0$ and $0 < S_0 < 1$. Write $s_n$ for the $n$-th partial sum. ($s_n = \sum_{k=0}^n (-1)^k a_k$) Now consider the new series $$S_1 = \sum_{n=0}^\infty (-1)^n \left ( \prod_{k = 1}^n s_k \right)$$ where we understand the empty product to be 1 (so that this is the first term.) This new series converges by the ratio test.

Can anything be said about its value? For example, if we always have $0 < S_1 < 1$, we could iterate the transformation to get a sequence of values $S_0, S_1, S_2, \dots$. Can anything be said about this sequence? I guess I'm wondering if anyone has seen this type of thing before (maybe it has a name?)
 http://mathoverflow.net/questions/34927/are-the-path-components-of-a-loop-space-homotopy-equivalent Are the path components of a loop space homotopy equivalent? Eric Finster 2010-08-08T14:24:25Z 2010-08-08T14:34:42Z
If $X$ is a based space, then we have $\pi_1(X) \cong \pi_0(\Omega X)$. This is to say we can identify elements in the fundamental group of $X$ with path components of the first loop space of $X$. My question is this: do all the path components of $\Omega X$ necessarily have the same homotopy type?

I ask because when you iterate this idea by considering $\pi_2(X) \cong \pi_1(\Omega X) \cong \pi_0(\Omega^2 X)$, the space $\Omega^2 X$ only sees the component of $\Omega X$ containing the basepoint (in this case the constant map). So when I imagine climbing up the tower of $X$'s homotopy groups, at each stage we apply the loop functor and discard all the information in the non-basepoint components.

If all the components of $\Omega X$ are homotopy equivalent, then this is clearly no loss of information, since we can identify the homotopy groups of $X$ with the (properly shifted) homotopy groups of any of the connected components. Otherwise, don't we generate much more information about $X$ by looking at the homotopy types of all the path components of $\Omega X$, $\Omega^2 X$, etc.?

I keep thinking there might be some argument about a certain map being a fibration, and this giving an equivalence between the various path components of $\Omega X$, realized as fibers of the map, or something like this, but I haven't found it yet, and I can think of no canonical maps which might serve as the desired equivalences.
 http://mathoverflow.net/questions/19084/equivalences-in-model-categories Equivalences in Model Categories Eric Finster 2010-03-23T03:57:40Z 2010-03-23T04:47:28Z
If $\mathcal M$ is a model category and I know that $A$ and $B$ are isomorphic in $\mathrm{Ho}(\mathcal M)$, is it guaranteed that there is a zig-zag of weak-equivalences in $\mathcal M$ connecting $A$ and $B$?
 http://mathoverflow.net/questions/17569/a-model-structure-on-symmetric-monoidal-categories A Model Structure on Symmetric Monoidal Categories Eric Finster 2010-03-09T03:21:08Z 2010-03-18T22:13:59Z
The recent article found here revisits Thomason's proof that symmetric monoidal categories model all connective spectra, but stops short of showing that there is a full closed model structure on this category (as does, it seems, Thomason's original paper.) Is there such a thing?

My guess is some lifting similar to how the model structure on small categories is derived would work, but I'm not sure if there are any complications.
 http://mathoverflow.net/questions/17425/homotopy-limits-over-fibered-categories Homotopy Limits over Fibered Categories Eric Finster 2010-03-07T22:35:27Z 2010-03-08T11:24:37Z
Suppose I have a small category $\mathcal{C}$ which is fibered over some category $\mathcal{I}$ in the categorical sense. That is, there is a functor $\pi : \mathcal{C} \rightarrow \mathcal{I}$ which is a fibration of categories. (One way to say this, I guess, is that $\mathcal{C}$ has a factorization system consisting of vertical arrows, i.e. the ones that $\pi$ sends to an identity arrow in $\mathcal{I}$, and horizontal arrows, which are the ones it does not. But there are many other characterizations.)

Now let $F : \mathcal{C} \rightarrow s\mathcal{S}$ be a diagram of simplicial sets indexed by $\mathcal{C}$. My question concerns the homotopy limit of $F$. Intuition tells me that there should be an equivalence

$$\varprojlim_{\mathcal{C}} F \simeq \varprojlim_{\mathcal{I}} \left (\varprojlim_{\mathcal{C}_i} F_i \right )$$

where I write $\mathcal{C}_i = \pi^{-1}(i)$ for any $i \in \mathcal{I}$, $F_i$ for the restriction of $F$ to $\mathcal{C}_i$ and $\varprojlim$ for the homotopy limit.

Intuitively this says that when $\mathcal{C}$ is fibered over $\mathcal{I}$, I can find the homotopy limit of a $\mathcal{C}$ diagram of spaces by first forming the homotopy limit of all the fibers, realizing that this collection has a natural $\mathcal{I}$ indexing, and then taking the homotopy limit of the resulting diagram.

Does anyone know of a result like this in the model category literature?

Update: After reading the responses, I was able to find a nice set of exercises here which go through this result in its homotopy colimit version.
 http://mathoverflow.net/questions/45951/sexy-vacuity/45993#45993 Comment by Eric Finster Eric Finster 2010-11-15T08:43:59Z 2010-11-15T08:43:59Z In a similar vein, Conway's theory of Games (which include the Surreal Numbers) gets quite a lot of mileage out of considering the "empty game" in which neither player has any move at all. Every game can be considered as a built from it, just as in Set theory. http://mathoverflow.net/questions/46083/controlling-reflective-subcategories-and-localizations Comment by Eric Finster Eric Finster 2010-11-14T23:28:52Z 2010-11-14T23:28:52Z Sorry, yes, this is a good point. I guess I really had the idea of "Bousfield Localizations" in my head, where one finds the desired category inside a given model category. http://mathoverflow.net/questions/41264/equivalence-of-cosimplicial-models-for-homotopy-pullbacks/41272#41272 Comment by Eric Finster Eric Finster 2010-10-08T12:48:18Z 2010-10-08T12:48:18Z Yes! I think I understand what you are saying better now. The difference between the two can be thought of as just a single subdivision of the path in $X_{12}$. So I wonder if the "canonical equivalence" you mention above can be actually realized as some kind of map induced by this subdivision. I'll give it some thought. Thanks for your help. http://mathoverflow.net/questions/41264/equivalence-of-cosimplicial-models-for-homotopy-pullbacks/41272#41272 Comment by Eric Finster Eric Finster 2010-10-06T14:09:14Z 2010-10-06T14:09:14Z Well, right. Except that somehow this explanation is not entirely satisfying to me. Shouldn't you expect a similar result in basically any (simplicial model) category? Spectra, for example? Or is this wrong? If it's true, then it leads me to believe there's some kind of "more general reason" that they are always equivalent. Maybe I am way off base here. http://mathoverflow.net/questions/41142/geometric-interpretation-of-universal-enveloping-algebras/41189#41189 Comment by Eric Finster Eric Finster 2010-10-05T18:25:23Z 2010-10-05T18:25:23Z I really like your first description in terms of distributions. Do you by chance have a reference for where I might learn about this? http://mathoverflow.net/questions/39901/set-theory-within-the-framework-of-category-theory Comment by Eric Finster Eric Finster 2010-09-24T22:55:15Z 2010-09-24T22:55:15Z While you ask specifically about "naive set theory" you might be interested in learning about the concept of a "topos." This is, roughly, a category which looks enough like the category of sets that you can think of it as a kind of generalized set theory. There is a vast literature on this subject: the books by Moerdijk and MacLane, Johnstone, and Goldblatt are three good places to start. http://mathoverflow.net/questions/38856/jokes-in-the-sense-of-littlewood-examples/39014#39014 Comment by Eric Finster Eric Finster 2010-09-24T16:00:01Z 2010-09-24T16:00:01Z I think the second author ought to be "Leinster," though I would love to take credit since I like that paper very much. :) http://mathoverflow.net/questions/38005/are-sieves-in-locally-small-categories-still-sets Comment by Eric Finster Eric Finster 2010-09-07T21:03:14Z 2010-09-07T21:03:14Z I think your observation about a category with a terminal object is correct. You could even that $C = \mathcal Set$ itself as an example. Do you have some objection to a sieve being a class of arrows closed under precomposition? http://mathoverflow.net/questions/37992/the-homotopy-cofiber-of-the-smash-product-of-two-maps-of-spectra/38002#38002 Comment by Eric Finster Eric Finster 2010-09-07T19:39:49Z 2010-09-07T19:39:49Z Yes, I somehow figured that a set formula was too much to ask for. But thanks for the tips. I will try these in my situation and see if I can get anything to fall out. http://mathoverflow.net/questions/37647/the-dold-thom-theorem-for-infinity-categories/37815#37815 Comment by Eric Finster Eric Finster 2010-09-05T21:02:19Z 2010-09-05T21:02:19Z I think maybe I see the descrepancy: to a space $X$ I want to associate some diagram $D(X)$ whose homotopy colimit is $SP^\infty X$. Regard this as a functor $D : \mathcal T \rightarrow \mathcal T^{\mathcal C}$ where $\mathcal C$ is the proposed indexing category. While it's true that homotopy colimits commute, there's no reason to suppose that the functor $D$ must preserve homotopy colimits. It can certainly be functorial without doing so (as is I suspect the case for $Q$.) Somebody please correct me if I am wrong about this. http://mathoverflow.net/questions/37647/the-dold-thom-theorem-for-infinity-categories Comment by Eric Finster Eric Finster 2010-09-05T19:27:57Z 2010-09-05T19:27:57Z @Dan: In the form I've stated it, yes, I suspect you are right (and I was using that paper as a reference when I wrote this.) On the other hand, there are many precursors which say that this should already be the right answer. I'm thinking about the Baratt-Kahn-Priddy theorem, Thomason's work on permutative categories, etc. http://mathoverflow.net/questions/37647/the-dold-thom-theorem-for-infinity-categories/37815#37815 Comment by Eric Finster Eric Finster 2010-09-05T19:14:40Z 2010-09-05T19:14:40Z By this argument, then, the functor $Q(-)$ should preserve homotopy pushouts. Is this true? In your particular example, I think the resulting square is definitely a homotopy pullback, since $\Omega Q(S^2) \simeq Q(S^1)$, but I'm unsure if it is also a homotopy pushout. Am I missing something? http://mathoverflow.net/questions/37647/the-dold-thom-theorem-for-infinity-categories Comment by Eric Finster Eric Finster 2010-09-04T09:02:54Z 2010-09-04T09:02:54Z @André: After mulling it over a bit last night, I came to a similar conclusion about somehow getting orbit categories into the picture, so this seems like a promising thing to look at. I'll report back if I find anything. http://mathoverflow.net/questions/30396/derived-algebraic-geometry-and-chow-rings-chow-motives Comment by Eric Finster Eric Finster 2010-08-22T14:40:03Z 2010-08-22T14:40:03Z I gather this is part of the motivation behind Jacob Lurie's paper on "Structured Spaces" (which you can find on his website) though I have not understood enough of it yet to see if he carries out this particular application. http://mathoverflow.net/questions/36016/understanding-the-etale-space-construction-from-a-formal-viewpoint/36028#36028 Comment by Eric Finster Eric Finster 2010-08-21T15:02:18Z 2010-08-21T15:02:18Z @David: Yes, this had occurred to me. It shouldn't change the proof much to consider all restrictions compatible with the given cover.


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## Minimum distance for echo

Class x > Physics > Sound

If d is the distance between the observer and the obstacle and V is the speed of sound, then the time taken for echo

t = $\frac { total\quad distance\quad travelled }{ Speed\quad of\quad sound }$ = $\frac { 2d }{ V }$

or, d =$\frac{Vt}{2}$

Now putting t = 0.1 s and V = 340 m/s, we get

d = $\frac { 340\times 0.1 }{ 2 }$ = 17 m.

Thus, to hear an echo, a minimum distance should be 17 m.

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# Cyclic heat engine

#### Da-Force

Isothermal? Whoa! Are you looking at the same picture I'm looking at?

The graph is Pressure versus Volume.

It's a horizontal line on the graph for 1-2 and 3-4 so it IS isobaric... *rubs eyes* Yeah, thats what it looks like.

#### Cyrus

Yes, it is Isobaric and it is isothermal (in this case). Go read your book about how a carnot cycle works, what does it tell you for process 1-2?

#### Da-Force

Now whats confusing me is that we are looking at a cyclic process and asked to apply it to a Carnot efficiency.

Now, if I understand correctly from my book, isobaric does not mean isothermal. I know that isotherms are always curved on a P versus T graph so if I proceed to draw isotherms....

http://xs73.xs.to/pics/06121/aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa.png [Broken]

So basically, I don't understand how you achieve isothermal in any case...

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#### endeavor

I know what a Carnot engine is. Two adiabatic processes plus two isothermal.
Well, the question says it's isobaric and isometric. But then it says, "What would be the efficiency of a Carnot engine operating with the same high-temperature and low-temperature reservoirs?" Does this mean that we are to assume that the graph is the Carnot engine? or that the low and high temperature reservoirs of the graph are that of the Carnot engine?
If it is the former, then how would I determine which processes are the adiabatic and which are the isothermal?

#### Cyrus

I thought your graph has 1-2 labeled 650K. That would make all values of T from 1-2 = 650k.

#### Da-Force

LOL! Well, I thought that too... But then it would have to have been curved... I know you CAN calculate at any point the T because that is definitely a given on any Pressure versus Volume graph... So tell me, did I explain that part correctly now that we've cleared up the confusion?

I got it! BOOOO

#### Cyrus

I thought your graph said 1-2 was a constant temperature line!

<Pulls my hair out>

None of it is constant temperature. You have to find the gas constant using:

$$PV=mRT$$

Then do as you guys tried by going around the entire process.

You will find that $$T_H = 835.714$$

and $$T_L = 390$$

use that in carnot and you will get:

$$\eta_{th,rev} = 1- \frac{T_L}{T_H} = 1 - .4666 = .5333$$

That was an AWFUL problem in my opinion.

#### Da-Force

cyrusabdollahi said:
I got it! BOOOO
Oh darn it :-(

Now this is the part where you explain how you did it with your ever-knowing knowledge and mind.

#### Da-Force

Oh wait.. Now I forget something... T(h) would be the hottest temperature and T(c) would be the coldest?

I kept thinking initial and final Explain it to me again

#### Cyrus

No, I just stole what you did. (with some refining) :rofl:

Last edited:

#### endeavor

Wow, thanks a lot!
I kept taking the inital temperature as the hot too!

#### Cyrus

Tell your teacher that problem is unclear.

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Matching with a Preamplifier
Resistance, capacitance, cables, oh my!
— How do I set capacitive and resistive loading for a Moving Magnet (MM) cartridge?
— There are a few "rules of thumb" that only apply to Moving Magnet and other high inductance cartriges (MI, fluxvalve, etc):

• The higher you go in resistive and capacitive loading, the "brighter" the sound is
• If the sound is too dull and lacks sparkle, increase resistive loading
• If the sound is too bright and tiring, decrease capacitive loading if possible. If not, decrease the resistive load
• No load adjustment is needed for modern MM-type cartridges (Ortofon OM/2M, Audio Technica VM, Sumiko and other Excel derivatives, Goldring). They are designed to be linear into a "standard" 47 kOhm/200 pF (picoFarad) load
• Typical tonearm cable has a 150-200 pF capacitance. Do not add more capacitive loading on the preamp side unless needed

Unlike the moving coil cartridges, there is no easy way to set the loading correctly. It will always be your subjective preference, unless you have a good test LP and a signal analyzer.

If you have the same recording on the LP and in digital format, and are happy with digital's tonal balance, you can try and match it by adjusting the loading. Digital releases are mastered differently, so there will never be an exact match.

Before the industry settled on today's typical 47 kOhm/200 pF loading, there existed considerable variation. In the 1970s, 75 kOhm loading was considered standard by some hi-fi manufacturers, and that's what the old Shures, Stantons and Pickerings might be expecting. If you have a vintage cartridge and it sounds dull, raise the resistive loading.

Many (but not all) 1970s, 80s and 90s Audio Technica cartridges are too bright into 47 kOhm. Loading them into 35 kOhm will result in a more linear response and a less fatiguing sound.

— How do I set capacitive and resistive loading for a Moving Coil (MC) cartridge?
— In general, setting the loading for the moving coil (MC) cartridge is a lot easier than doing the same for MM.

For impedance matching (resistive loading), start with the lowest available with your preamp or step-up transformer. Select a track you know well, with prominent highs and clean but full bass. Start increasing the loading, and listen to the track after each step.

Once you hear no difference in highs and bass with the increase in resistive loading, decrease it by one step and leave it there. That's it, you're done.
This method might not work for some MC cartridges. "High output" moving coils in particular need "standard" 47 kOhm loading.

— Why does capacitive and resistive loading of MM cartridge matter so much?
— The short answer: because of immense inductance of fixed coil cartridges (moving magnet, moving iron etc). 0.3 to 0.7 Henry is typical, more than 1 whole Henry is not unheard of. Paired with a huge inductance, even minuscule capacitances (50-200 pF) suddenly matter a lot.

— The long (but still somewhat simplified) answer:
Here is the electrical model of a typical fixed coil (MM/MI) cartridge, with inductance of 0.5 Henry and internal resistance of 1 kOhm. It is loaded into 200 picoFarad of capacitance and 47 kOhm of resistance.

The cartridge's inductance and the load capacitance form a resonant circuit. Because the coil inductance is so high, the resonance frequency $$f$$ is in the audible range. Depending on the cartridge, it can be as low as 12 kHz.

$$f={1\over{2\pi\sqrt{L_{cartridge}C_{load}}}}$$

The loading resistor $$R_{load}$$ is there to dampen the peak of this resonance. What would happen if we change its value?
The higher it is, the more pronounced the high frequency peak is. The curves above are shown for 200 picoFarad capacitive loading.

Changing the capacitive loading, we see that with higher capacitance, the peak becomes larger and lower in frequency. But also the subsequent high frequency roll-off becomes steeper. This is why, for subjective adjustment, it is better to change the loading resistance and not the capacitance. With resistance, only the HF peak amplitude changes. With capacitance, you change at once the peak frequency, amplitude, and high frequency roll-off.


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Browse Questions

Solve the given inequality graphically in two dimensional plane . $2x -3y > 6$

Toolbox:
• To represent the solution of linear inequality of one or two variable in a plane if the inequality involves $'\geq'$ or $' \leq$ we draw the graph of the line as a thick line to indicate the line is included in the solution set.
• If the inequality involves $'>'$ as $'<'$ we draw the graph of the line using is not included in the solution set.
• To solve an inequality $ax+by > c \qquad a \neq 0, b \neq 0 ( or \;> )$
• We consider the corresponding equation $ax+by =c$ which represents a straight line This line divides the plane into two half planes I and II
• We take any point in I half plane and check if it satisfies the given inequality will be one half plane (called solution region ) Containing the point satisfying the inequality
Step 1:
The given inequality is $2x -3y >6$
Consider the equation $2x-3y=6$
We see taht the points $(3,0)$ and $(0,-2)$ satisfy the equation.
The graphical representation of the line is given below.
Step 2:
The line divides the XY -plane into two half- plane I and II .
Consider a point (0,0) in half plane I
We see that $2(0) -3(0) >6$
$0 > 6$ is false
The half plane I containing the point (0,0) is not the solution region of the inequality.
Step 3:
Thus the solution region of the given inequality is the half plane II excluding the line . It is represented by the shaded region.


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# Second isomorphism theorem

Algebra Level 2

Let $S_3$ be the symmetric group on three letters. Let $H$ be the subgroup generated by the transposition $(12),$ and let $N$ be the subgroup generated by the transposition $(13).$

Then the second isomorphism theorem gives an isomorphism $HN/N \simeq H/(H \cap N),$ where $HN = \{ hn : h \in H, n \in N\}.$

Now $H \cap N = \{1\},$ so the right side has order $2.$ So the left side has order $2.$ Now $|N|=2$ and $|HN/N|=2,$ so $|HN|=4.$ But Lagrange's theorem says that $|HN|$ must divide $|S_3|,$ which is $6.$

What has gone wrong with this argument?

I. As defined above, $HN$ is not a subgroup of $S_3,$ so Lagrange's theorem does not apply.
II. The order of a quotient $G/K$ of finite groups is not always equal to $|G|/|K|.$
III. $N$ is not normal, so the second isomorphism theorem does not apply.

×


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# Confusion with bilinear form and linear operation relation.

My lecture notes say that for every bilinear form there exists a linear operator such that $$\tau (v,w) = v.(Tw)$$ and that there must exist some other linear operator $S$ such that $$(Sv).w = v.(Tw).$$ I understand everything up to there but then it says that it's easy to see that in an orthonormal basis, the matrix of S is just the transpose of the matrix of T. I can't get my head around why it has to be an orthonormal basis. Surely, if $A$ is the matrix for $T$ and $B$ is the matrix for $S$ then, $$(Sv).w = v.(Tw)$$ $$(B\underline{v})^T\underline{w} = \underline{v}^T A\underline{w}$$ $$\underline{v}^T B^T \underline{w} = \underline{v}^T A\underline{w}$$ So $B^T = A$ for any basis? Where am I going wrong?

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How come that your "dot product" on the original vectorspace is (in your basis) the euclidian scalarproduct in $\mathbb R^n$? –  Blah Apr 6 '12 at 15:28
Because the standard scalar product $\underline{x}.\underline{y} =\underline{x}^T \underline{y}$ so if A is the matrix of the bilinear form, $\tau(v,w) = \underline{v}^T A \underline{w} = v.(Tw)$? –  user26069 Apr 6 '12 at 16:15
could you please describe the situation? Is your vectorspace $\mathbb R^n$ and "." the euclidian scalarproduct? –  Blah Apr 6 '12 at 16:45

By using that $v.\omega$ is $\underline{v}^T\underline{w}$, you are assuming that your basis is orthonormal. Because if, say, $V$ is 2-dimensional and $f_1,f_2$ is your basis, then you seem to claim that you see $f_1$ as $[1\ 0]^T$, $f_2=[0\ 1]^T$, and that $$f_1.f_2=([1\ 0]^T)^T[0\ 1]^T=0.$$

That's of course not necessarily the case. Consider $\mathbb{R}^2$ with the basis $f_1=(1,0)$, $f_2=(1,1)$, with the usual dot product. Then, as vectors in the basis $\{f_1,f_2\}$, we associate $f_1$ with $[1\ 0]^T$ and $f_2$ with $[0\ 1]^T$, but $f_1.f_2=1$, which cannot be obtained by doing the product $f_1^Tf_2$ as matrices in their own basis.

Still in this basis, consider the operator $T$ given by $Tf_1=f_2$, $Tf_2=0$. Then, in the basis $\{f_1,f_2\}$, the operator $T$ is represented by the matrix $$\begin{bmatrix}0&1\\ 0&0\end{bmatrix}.$$ Let us calculate the entries of $S$ (usually denoted by $T^*$ and called the adjoint of $T$) in the basis $\{f_1,f_2\}$: \begin{eqnarray} S_{11}&=&(Sf_1).f_1=f_1.(Tf_1)=f_1.f_2=1,\ S_{12}&=&(Sf_2).f_1=f_2.Tf_1=f_2.f_2=2,\ S_{21}&=&(Sf_1).f_2=f_1.Tf_2=f_1.0=0,\ S_{22}&=&(Sf_2).f_2=f_2.Tf_2=f_2.0=0. \end{eqnarray} So, in the basis $\{f_1,f_2\}$, the matrix for $S$ is $$\begin{bmatrix} 1&2\\ 0&0 \end{bmatrix}$$

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There is nothing wrong, and the result is true for any basis.

However, the analysis above skips over the tedious bits by jumping from the vectors ($v, w, S v, T w$) in the vector space $V$ to their representations in $\mathbb{R^n}$ ($\underline{v}, \underline{w}, B \underline{v}, A \underline{w}$). This tedium is considerably reduced by choosing an orthonormal basis.

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My answer is a little glib; to obtain the matrix representation of $T$ or $S$, a linear bijection $L:\mathbb{R^n} \rightarrow V$ is needed. However, you cannot use just any map, you need to preserve the dot (inner) product, so that if $x,y \in \mathbb{R^n}$, then $Lx . Ly = x^T y$. –  copper.hat Apr 6 '12 at 23:36


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# Tag Info

Hint Let $\,H_0(x,t)=1$ , $H_1(x,t)=x$ and for every $n\ge 2$ set $${{H}_{n}}(x,t)=x {{H}_{n -1}}(x,t)-(n-1)\,t\,{{H}_{n-2}}(x,t)$$ then ${{H}_{n }}(W_t ,t)$ is a Martingale. For exapmple $$H_1(W_t,t)=W_t$$ $$\qquad H_2(W_t,t)=W_t^2-t$$ $$\qquad\qquad H_3(W_t,t)=W_t^3-3tW_t$$ $$\vdots$$