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In Lurie's "On the Classification of Topological Field Theories," he states in Proposition 1.1.8 that for an oriented compact manifold $M$ and a TQFT $Z:\mathrm{Cob}(n)\to \mathrm{Vect}_k$, there is a perfect pairing $Z(\overline{M})\otimes Z(M)\to k$ where $\overline{M}$ denotes $M$ with the opposite orientation. In the proof of this, he mentions the easily defined map $\alpha:Z(\overline{M})\to Z(M)^\vee$, where $^\vee$ denotes the dual space, which relies on the evaluation map coming from the cobordism $M\coprod\overline{M}\to\emptyset$ associated to $M\times[0,1]$. He intends to describe its inverse $\beta:Z(M)^\vee\to Z(M)^\vee\otimes Z(M)\otimes Z(\overline{M})\to Z(\overline{M})$ as a composite of the coevaluation map from $\emptyset\to M\coprod\overline{M}$ (i.e. $k\to Z(M)\otimes Z(\overline{M})$) followed by the bilinear pairing of $Z(M)$ with its dual. He then says that by "judiciously applying the axioms for a topological field theory, one can deduce that $\beta$ is an inverse to $\alpha$."

I am having trouble seeing that this is so. Just looking for maybe a little help, not necessarily the entire proof here, but anything would be appreciated? Is it futile to look at elements of the vector spaces considering that we don't really know much about what's going on there? Is this an entirely categorical proof, and if so, which "axioms" should I be looking the most closely at, the coherence ones for tensor functors?

Thanks!

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I'm not sure we need those two tags [tft] and [tqft], if you do think they are needed please start a meta thread explaining why. Either way, I don't think these acronyms are known enough to be used widely so even if you do end up using those tags please use full words. –  Asaf Karagila Jan 9 '12 at 20:16
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Sure, no need for both tags. However, one would be nice, and using full words causes there to be too many characters. You decide, it's an active area of research... I'm a little surprised at your thoughtless deletion though. –  Jon Beardsley Jan 9 '12 at 20:20
    
I'm sure that there are many active research areas as this one, however the consideration is what is the expected volume for questions in the topic. As for the abbreviations, if [topological-vector-spaces] is a tag, I don't see TFT causing much difficulty. –  Asaf Karagila Jan 9 '12 at 20:26
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This is vague and I should actually read what you wrote and then I'll answer, but in the meantime you should look at the picture on page 6: math.ucsd.edu/~mshulman/papers/traces_sym.pdf –  Dylan Wilson Jan 10 '12 at 0:37
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(I meant that my answer was vague, not your question. And I think the reason you get a true dual is that the "triangular identities" you need to check correspond to showing that some twisted up manifold is diffeomorphic to the untwisted up manifold... in particular for like the 1-dimensional case I think you can literally use the picture on page 6 of the paper. I'm probably misunderstanding something though.) –  Dylan Wilson Jan 10 '12 at 19:44

1 Answer 1

up vote 3 down vote accepted

Okay, so first a little notation. Let $V=Z(M)$, $V'=Z(\overline{M})$ and $V^\vee=Hom(V,k)$, i.e. the dual space of $V$. Now, we want to basically show an isomorphism between $V'$ and $V^\vee$.

Let's recall that we have the following maps in $\mathrm{Cob}(M)$:

$e:M\coprod\overline{M}\to\emptyset$

$c:\emptyset\to\overline{M}\coprod M$

$1_M:M\to M\coprod\overline{M}\coprod M\to M$ and

$1_\overline{M}:\overline{M}\to \overline{M}\coprod M\coprod \overline{M}\to \overline{M}$

where the latter two maps are compositions of $c$ and $e$ and are equal to the identity by the text Dylan referenced (i.e. sort of S shaped manifolds, or Zorro's lemma?).

From these morphisms we get linear maps in $\mathrm{Vect}_k$ (I will still just use $c$ and $e$ and the category will be clear from context):

$e:V\otimes V'\to k$

$c:k\to V\otimes V'$

$1_V:V\to V\otimes V'\otimes V\to V$

$1_{V'}:V'\to V'\otimes V\otimes V'\to V'$.

By $eval$ and $coev$ we denote the standard evaluation and coevaluation maps associated to $V$ and its dual. Recall that these maps also satisfy triangle inequalities similar to above.

Since $e$ defines a bilinear pairing on $V\otimes V'$, we can define $\alpha:V'\to V^\vee$ such that $v'\mapsto e(-,v')$.

Next we can define a map $\beta=(1_{V^\vee}\otimes c)\circ (eval\otimes 1_{V'}):V^\vee\to V^\vee\otimes V\otimes V'\to V'.$ We show that $\beta\circ\alpha=1_{V'}$ and $\alpha\circ\beta=1_{V^\vee}$. This follows from the following diagrams, which are easily seen to be commutative:

$V'\overset{\alpha}\longrightarrow V^\vee\overset{1_{V^\vee}\otimes c}\longrightarrow V^\vee\otimes V\otimes V'\overset{eval\otimes 1_{V'}}\longrightarrow V'$

$\downarrow ~~~~~~~~~~~~~~~~~~~~~~~~~~~~\uparrow_{\alpha\otimes 1_{V}\otimes 1_{V'}}~~~~~\uparrow$

$\searrow\overset{1_{V'}\otimes c}\longrightarrow\longrightarrow V'\otimes V\otimes V'\overset{e\otimes 1_{V'}}\longrightarrow\nearrow$

$V^\vee\overset{1_{V^\vee}\otimes c}\longrightarrow V^\vee\otimes V\otimes V'\overset{eval\otimes 1_{V'}}\longrightarrow V'\overset{\alpha}\longrightarrow V^\vee$

$\downarrow ~~~~~~~~~~~~~~~~~~~~~~\downarrow_{1_{V^\vee}\otimes 1_{V}\otimes\alpha}~~~~~~~~~~~~~~\uparrow$

$\searrow\overset{1_{V^\vee}\otimes coev}\longrightarrow V^\vee\otimes V\otimes V^\vee\overset{eval\otimes 1_{V^\vee}}\longrightarrow\nearrow$

Sorry for the horrific typesetting.

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