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Classical formulation

Consider the (untwisted) Dijkgraaf-Witten invariant, defined for an oriented, connected, closed manifold $M$ and a finite group $G$:

$$DW_G(M) := \lvert \operatorname{Hom}(\pi_1(M), G )\rvert $$

We're counting group homomorphisms from the fundamental group of the manifold to $G$. I am multiplying the more common definition by the size of the group, in order to simplify the following formulas. (Some are of the opinion that each homomorphism should be weighted by the inverse size of its automorphism group. I think this is not so, see e.g. this article. Instead, I think you can also sum over equivalence classes of $G$-bundles, which correspond to conjugacy classes of homeomorphisms, and there you need the weights.)

Now this invariant satisfies a simple identity, where $\#$ denotes connected sum:

$$DW_G(M_1 \# M_2) = \lvert \operatorname{Hom}(\pi_1(M_1 \# M_2), G )\rvert = \lvert \operatorname{Hom}(\pi_1(M_1) * \pi_1(M_2), G )\rvert = \lvert \operatorname{Hom}(\pi_1(M_1)) \rvert \cdot \lvert \operatorname{Hom} (\pi_1(M_2), G )\rvert = DW_G(M_1) \cdot DW_G(M_2) $$

The Dijkgraaf-Witten invariant is multiplicative under direct sum.

We've used the free product of groups, denoted as $*$. We know from the Seifert-van-Kampen theorem that $\pi_1(M_1 \# M_2) = \pi_1(M_1) * \pi_1(M_2)$.

Homotopy theory formulation

We can also write the Dijkgraaf-Witten invariant as the number of homotopy classes into the classifying space of $G$:

$$DW_G(M) = \lvert [M, BG] \rvert$$

Indeed, instead of $BG$ we could insert any (sufficiently finite) homotopy 1-type. This calls for a generalisation, known as the Yetter invariant.

An aside: The twisted invariant

Let $[M]$ be the fundamental class of the oriented manifold $M$, $\omega \in H^{\operatorname{dim} M}(G, k^\times)$ a group cohomology element (with values in the unit group of a field) and $c\colon M \to B(\pi_1(M))$ the canonical map. Then the twisted Dijkgraaf-Witten invariant is defined as:

$$DW_G^\omega(M) := \sum_{\phi\colon \pi_1(M) \to G} \langle \phi^*(\omega), c_*([M]) \rangle $$

The inner product $\langle - , - \rangle$ comes from Poincaré duality.

It is easy to see that if $\omega$ is the trivial cocycle, the sum ranges over $1$, and we recover the original formula.

Yetter invariant

David Yetter, and then later Tim Porter, defined, for a homotopy 2-type $\mathcal{T}$, the following invariant:

$$Y_\mathcal{T}(M) = \lvert [M, \mathcal{T}] \rvert$$

I think the original definition is more hands-on, in terms of crossed modules.

Edit: I seem to have omitted the correct groupoid cardinalities here. See Arun Debray's answer further below for the correct version, or this article.

Since there are some kinds of higher formulations of Seifert-van-Kampen (as discussed in this question) suitable for this situation, I'm asking the following question:

Question: Is the Yetter invariant multiplicative under connected sum? I.e. does the following hold:

$$Y_\mathcal{T}(M_1 \# M_2) = Y_\mathcal{T}(M_1) \cdot Y_\mathcal{T}(M_2)$$

Manuel Bärenz
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    I'm puzzled by your characterization of the (untwisted) DW invariant. It should be the sum, over $\rho : \pi_1(M)\to G$, of $1/|Stab(\rho)|$. For example, the trivial representation contributes $1/|G|$, not 1. – Kevin Walker Mar 14 '18 at 18:01
  • @KevinWalker, I think what you're referring to is the definition of the state spaces of Dijkgraaf-Witten. It's spanned by conjugation equivalence classes of connections on the boundary. But I'm fairly certain the state sum is over all homomorphisms, without that factor. – Manuel Bärenz Mar 14 '18 at 18:55
  • @ManuelBärenz how does one weight a vector space by $1/|\mathrm{Stab}(\rho)|$? I also thought the weighting was for partition functions (i.e. Dijkgraaf-Witten invariants), rather than state spaces. – Arun Debray Mar 14 '18 at 19:13
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    $\ast$ is called the free product, not the direct product. – Theo Johnson-Freyd Mar 14 '18 at 21:16
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    @Kevin: I think OP's "DW invariant" of $M$ is your DW invariant of $M \times S^1$. – Theo Johnson-Freyd Mar 14 '18 at 21:17
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    Your invariant is not multiplicative when $\dim M = 1$. – Theo Johnson-Freyd Mar 14 '18 at 21:22
  • @KevinWalker, have a look at this answer. If you sum over $G$-bundles weighted by (the inverse size of) their automorphism group, you get the formula I wrote, divided by $\lvert G \rvert$. – Manuel Bärenz Mar 15 '18 at 11:07
  • @TheoJohnson-Freyd, yes, of course it's the free product, thanks! – Manuel Bärenz Mar 15 '18 at 11:12
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    @ArunDebray, I meant that weighting by the inverse size of a group categorifies to taking equivalence classes with respect to the group action. – Manuel Bärenz Mar 15 '18 at 11:34
  • @KevinWalker, for another example from the literature, consider proposition 4.4 from Cui's article about $G$-crossed state sum models. – Manuel Bärenz Mar 15 '18 at 11:41
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    @ManuelBärenz, I see. You are using homomorphisms rather than conjugacy classes of homomorphisms, so this is a renormalized (by $|G|$) version of the DW TQFT path integral. I was confused since most people (including Dijkgraaf and Witten) use the TQFT normalization. – Kevin Walker Mar 15 '18 at 13:20
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    @TheoJohnson-Freyd, I think your claim about this being the DW TQFT invariant for $M\times S^1$ would be correct if the OP were counting (unweighted) conjugacy classes, but he is counting homomorphisms, so I think the $M\times S^1$ claim is not correct. – Kevin Walker Mar 15 '18 at 13:23
  • @KevinWalker, yes, apologies for the confusion. Although unconventional, it simplifies the derivation of the connected-sum formula. – Manuel Bärenz Mar 15 '18 at 14:26
  • @KevinWalker You are correct. I’m not sure where my confusion came from. I suspect I was tired and confused. – Theo Johnson-Freyd Mar 16 '18 at 04:56

1 Answers1

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As Kevin Walker pointed out in a comment, Dijkgraaf-Witten invariants are weighted by $1/\mathrm{Stab}(\rho)$. In the same way, the Yetter invariant for $\mathcal T$ and $M$ is generally defined such that it's weighted using the 2-groupoid cardinality of $\pi_{\le 2}\mathrm{Map}(M, \mathcal T)$, so that the invariant is $$\sum_{[f\colon M\to\mathcal T]} \frac{|\pi_2(\mathrm{Map}(M, \mathcal T), f)|}{|\pi_1(\mathrm{Map}(M, \mathcal T), f)|}.$$

If we use this normalization, the Yetter invariants are the partition functions of a TQFT $Z_{\mathcal T}$, usually called the Yetter model. In this case, a different MathOverflow answer by Kevin Walker tells us that $Z_{\mathcal T}$ is multiplicative under connect sum iff

  • $\dim Z_{\mathcal T}(S^{n-1}) = 1$, and
  • $Z_{\mathcal T}(S^n) = 1$.

The state space $Z_{\mathcal T}(M^{n-1}) := \mathbb C[[M, \mathcal T]]$, so for $n = 3$, the first property doesn't hold: $\dim Z_{\mathcal T}(S^2) = |\pi_2(\mathcal T)|$. A similar problem occurs for $n = 2$.

If $n > 3$, then $[S^{n-1}, \mathcal T] = 0$, so the first property holds. The second property does not quite hold: $[S^n, \mathcal T] = 0$, but we have to calculate the weighting. Since $\mathrm{Map}(S^n, \mathcal T)\simeq\mathrm{Map}(\mathrm{pt}, \mathcal T)\cong \mathcal T$, $$Z_{\mathcal T}(S^n) = \frac{|\pi_2(\mathcal T)|}{|\pi_1(\mathcal T)|},$$ which is frequently not equal to 1.

I don't know about the unweighted version you mentioned, since it doesn't come from a TQFT as far as I know.

Arun Debray
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    As I said in the comments, I think OP's weightings are your waitings for $M \times S^1$. Reduction along $S^1$ takes TFTs to TFTs. – Theo Johnson-Freyd Mar 14 '18 at 21:19
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    What you've explained is that the answer to OP's question is "yes" if certain homotopy groups of $\mathcal T$ vanish. Which homotopy groups? It depends on the dimension of $n$. – Theo Johnson-Freyd Mar 14 '18 at 21:20
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    In particular, note that OP's invariant is not multiplicative for connect sums when $\dim M = 1$. Then $S^1 # S^1 = S^1$, and $Z(S^1) =$ number of conjugacy classes in $G$, which is typically nonzero. – Theo Johnson-Freyd Mar 14 '18 at 21:21
  • I think this is a misunderstanding. The weights only need to show up in a different formulation of Dijkgraaf-Witten. See e.g. this article. – Manuel Bärenz Mar 15 '18 at 11:20
  • Oh, I see, but if we multiply Yetter by $Z_\mathcal{T}(S^n)$, we arrive at a multiplicative invariant once again! (Or, as per Kevin's suggestion, with an Euler characteristic theory if $n$ is even) – Manuel Bärenz Mar 15 '18 at 14:49
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    In the comments to the original post, Kevin Walker corrected my mistake. This is not the reduction of DW theory along S^1. It is simply |G| times the DW invariant. – Theo Johnson-Freyd Mar 16 '18 at 04:57