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• CommentRowNumber1.
• CommentAuthorTim_Porter
• CommentTimeDec 11th 2009
I have made a comment on the groupoid cardinality page. It draws peoples attention to Quinn's notes on TQFTs which uses a notion of homotopy order to construct scaling factors in a simple TQFT. This does not seem to be mentioned in stuff that I have seen and googling gets very few hits for this. It is clearly the same as groupoid cardinality when both are defined.
1. there’s something puzzling me about groupoid cardinality. namely, if $X$ is a tame groupoid and $f:X\to\mathbb{C}$ a function, then we have a good prescription for the integral of $f$ over $X$. since, as recalled at n-vector space, the set $\mathbb{C}$ is identified wit te 0-category $0Vect_\mathbb{C}$, this means that we know how to integrate a functor $X\to 0Vect_\mathbb{C}$.

but what if we move from $0Vect_\mathbb{C}$ to $1Vect_\mathbb{C}$? in this case we have a tentative answer given by sections of the vector bundle corresponding to $f:X\to 1Vect_\mathbb{C}$, but this is a satisfactory answer only in case $X$ is a 1-type. in general, I feel the vector space of sections alone forgets about the higher homotopy of $X$. this is upsetting, since the integral of a $0Vect_\mathbb{C}$ remembers higher homotopy, so taking sections seems to break the pattern.

however even in the 1-type case, things are not as neat as they could be. namely, assume $X$ to be a conected 1-type, so that $X=\mathbf{B}G$ for some group $G$. Then the datum of $f:X\to 1Vect_\mathbb{C}$ is the datum of a linear representation $V$ of $G$, and taking sections corresponds to taking the subspace $V^G$ of $G$-invariants. but this is only the tip of the iceberg giving cohomology of $G$ with coefficients in the $G$-module $V$. so maybe one should think of the integral of a functor $X\to 1Vect_\mathbb{C}$ as the simplicial vector space whose associated complex computes the group cohomology of $G$ with coefficients in $V$. have to think more on this.

• CommentRowNumber3.
• CommentAuthorUrs
• CommentTimeMay 13th 2010
• (edited May 13th 2010)

I feel the vector space of sections alone forgets about the higher homotopy of $X$. this is upsetting, since the integral of a $0Vect_\mathbb{C}$ remembers higher homotopy, so taking sections seems to break the pattern.

Yes, I know what you mean. Likely what we need is not the $n$-category of $n$-vector spaces, but the $(\infty,n)$-category of $(\infty,n)$-vector spaces.

The extra $\infty$ in there will make sure that everything always depends on all higher homotopies.

It’s not too hard to make some guesses here:

Set

$(\infty,1)Vect_k := Ch_\bullet(k)$

to be the $(\infty,1)$-category of chain complexes. This is still symmetric monoidal, so we can base the iterative definition

$(\infty,n)Vect := (\infty,n-1)Vect Mod$

on that.

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeMay 13th 2010
• (edited May 13th 2010)

And maybe better, what we should do is:

take to be $k$ be a commutative simplicial ring. Then $(\infty,1)Vect := k Mod$ its $(\infty,1)$-category of simplicial modules, and then iterate.

This would also serve to “categorfy” the ground field itself. Which is now an $\infty$-groupoid with ring structure, i.e. really a special case of an $E_\infty$-spectrum.

Right, so probably in full generality we should go this way:

• fix $k$ an $E_\infty$-ring spectrum.

• declare $(\infty,0)Vect_k := k$ – a symmetric monoidal $\infty$-groupoid.

• then iterate: $(\infty,n)Vect_k := (\infty,n-1)Vect_k Mod$.

2. sounds good. that would fit what I was writing, the comment in section “Ch(Vect)-enriched categories” in 2-vector space and, most remarkably, the fact that the natural target in the cobordism hypothesis is a symmetric monoidal $(\infty,n)$-category. now, to have a pattern, we would like that in building the complex of sections on a (oo,1)-vector bundle, a weighted sum appears. let’s see what happens in the simplest case: a finite group representation, seen as a morphism $f:\mathbf{B}G\to 1Vect_k\hookrightarrow (\infty,1)Vect_k$. let $V$ be the vector space associated with the unique object of $\mathbf{B}G$. then for every $g$ in $G$ we have an endomorphism $\rho_g:V\to V$ and $V^G$ is the subspace of $V$ where the operator $\frac{1}{|G|}\sum_{g\in G}\rho_g:V\to V$ acts as the identity. but it’s still not clear to me which the general receipt should be.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeMay 13th 2010
• (edited May 13th 2010)

Wait, once we embed all the way $f : \mathbf{B}G \to 1 Vect_k \hookrightarrow (\infty,1)Vect_k$ we also need to compute the limit/colimit of that functor there. This will no longer be just $V^G$, I think.

Ahm, let me see, what will it actually be. Er. Have to think more about that.

Notice the remarkable observation from FHLT, recorded at category algebra: if we compute the colimit of the trivial representation, but regarded as a representation on $2 Vect_k$

$\mathbf{B}G \stackrel{const_{1}}{\to} 2 Vect$

the result, which by definition is an algebra, is the group alghebra of $G$.

• CommentRowNumber7.
• CommentAuthordomenico_fiorenza
• CommentTimeMay 13th 2010
• (edited May 13th 2010)

we also need to compute the limit/colimit of that functor there

yes. we should obtain a complex whose $H^0$ is $V^G$. what I was trying to obtain was a bit of this in which a weighted sum of the $\rho_g$’s appeared. namely, the general picture should say not only “closed top dimensional manifolds go to numbers given by weighted sums, closed codimension one manifolds go to vector spaces (or complexes) and top dimensional manifolds with boundary go to linear maps between these vector spaces”, but also something like “matrix entries of these linear maps are given by weighted sums”. stating this correctly (and proving it) should be the key to understand the interplay between sections and integrals with the groupoid measure.

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeMay 13th 2010

Yes, right. I need to think…

3. a good exercise could be computing the dimension of the space of sections of a 1-vector bundle over a connected 2-type (e.g. realized as a crossed module). I’ll try to work out the details of this later.

4. still very confused about the above. need to think to something else for a few days before coming back to this.

• CommentRowNumber11.
• CommentAuthorDavid_Corfield
• CommentTimeMar 7th 2023

• Lior Yanovski, Homotopy Cardinality via Extrapolation of Morava-Euler Characteristics (arXiv:2303.02603)
• CommentRowNumber12.
• CommentAuthorGuest
• CommentTimeMar 7th 2023

what is the groupoid cardinality of the circle type?

• CommentRowNumber13.
• CommentAuthorUrs
• CommentTimeMar 7th 2023
• (edited Mar 7th 2023)

I forget how people define the groupoid cardinality for non-finite fundamental groups, if they do, but the evident naive answer comes out as expected:

$S^1 \simeq \mathbf{B}\mathbb{Z} \;\simeq\; \ast \sslash \mathbb{Z} \;\;\;\;\mapsto\;\;\;\; 1/\vert \mathbb{Z}\vert \;=\; 0$

(Though this may be a meaningless accident.)

• CommentRowNumber14.
• CommentAuthorDavid_Corfield
• CommentTimeMar 7th 2023

There was a funny idea of Dolan and Baez once that with the characteristic of the 2-sphere being $2$, there should be a weird convergence in the product of the homotopy cardinality.

• CommentRowNumber15.
• CommentAuthorUrs
• CommentTimeMar 7th 2023
• (edited Mar 7th 2023)
$S^2 \;\;\;\;\mapsto\;\;\;\; \frac{ {\vert \mathbb{Z} \vert} \cdot 2 \cdot 2 \cdot 3 \cdot 2 \cdot 24 \cdot 4 \cdots }{ {\vert \mathbb{Z} \vert} \cdot 2 \cdot 12 \cdot 2 \cdot 15 \cdot 4 \cdot 336 \cdots } \;\;\; \overset{???}{=} \;\;\; 2$
• CommentRowNumber16.
• CommentAuthorDavid_Corfield
• CommentTimeMar 7th 2023

Sounds crazy, doesn’t it? Perhaps some cohomological approximation, as in the Morava K-theories?

It’s even groups on top though so

$S^2 \;\;\;\;\mapsto\;\;\;\; \frac{ {\vert \mathbb{Z} \vert} \cdot 2 \cdot 12 \cdot 2 \cdot 15 \cdot 4 \cdot 336 \cdots }{ {\vert \mathbb{Z} \vert} \cdot 2 \cdot 2 \cdot 3 \cdot 2 \cdot 24 \cdot 4 \cdots } \;\;\; \overset{???}{=} \;\;\; 2$

The one good thing I noted at the time was that $S^3$ shares the same groups from $\pi_3$, so we should have $S^3 \mapsto 2/\vert \mathbb{Z} \vert = 0$, which is right.

• CommentRowNumber17.
• CommentAuthorUrs
• CommentTimeMar 7th 2023

One would need some notion of approximating general homotopy types by sequences of $\pi$-finite homotopy types.

• CommentRowNumber18.
• CommentAuthorGuest
• CommentTimeMar 7th 2023

What about $\mathbf{B} \mathbb{R}$? Is it’s cardinality also equal to zero, even though $|\mathbb{R}| \gt |\mathbb{Z}|$ as set cardinalities?

• CommentRowNumber19.
• CommentAuthorUrs
• CommentTimeMar 8th 2023

To say it more explicitly: A priori, groupoid cardinality is only defined for pi-finite homotopy types.

• CommentRowNumber20.
• CommentAuthorGuest
• CommentTimeMar 8th 2023

By the equivalence between cardinals and ordinals wouldn’t it make sense to move from the real numbers to the surreal numbers for groupoid cardinalities for homotopy types with infinite homotopy groups?

Theresa

• CommentRowNumber21.
• CommentAuthorGuest
• CommentTimeMar 8th 2023

Theresa,

That doesn’t work in the absence of the axiom of choice because one can only prove the equivalence of cardinals and ordinals in the axiom of choice

• CommentRowNumber22.
• CommentAuthorGuest
• CommentTimeMar 8th 2023

is the entire set of real numbers even needed in the definition of groupoid cardinality? I don’t think there are any infinity-groupoids with well-defined negative cardinality. One could probably do the same thing as in metric spaces and restrict the cardinality set to the non-negative real numbers.

• CommentRowNumber23.
• CommentAuthorUrs
• CommentTimeMar 29th 2023
• (edited Mar 29th 2023)

added DOI and publication year to this item:

and bruched-up the way it is referenced in the text.