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In Braided Monoidal Categories by Joyal and Street, §6 a monoidal category is $V =T(G,M,h)$ built using a recipe:

  • objects are are elements of $G$ ✓

  • $V_0(x,y) = M$ if $( x=y)$ or else $\varnothing$ if $(x\neq y)$.

    • is it a concidence it looks like the dirac-δ function ?
    • Can you get an identity like $\delta(x-y) = \sum e^{imx}e^{-iny} $ ?

  • composition is addition in $M$

    • below, should $x$ and $y$ be added or multiplied?

  • tensor product is $ (x \stackrel{\mu}{\to} x) \otimes (y \stackrel{\nu}{\to} y) \to (xy \stackrel{\mu\nu}{\to} xy) $

    • It's not clear to me what kind of object $(x \stackrel{\mu}{\to} x)$ should be ?
    • $\mu$ and $\nu$ are not defined ?
  • the associativity morphism is $h(x,y,z) = (xy)z \to x(yz)$ ✓
  • the identity element of $G$ acts a a strict identity element of $V$ ✓

My main question is how the associativity of $G$ is getting twisted by the 3-cocycle ? All my other questions on the page are subordinate to this one.

The examples I have in mind come from specific choices of $G$ and modules $H$ for discussion on a future date.

All I know about group cohomology is that it generalizes Carries in arithmetic:

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  • $\begingroup$ $\mu$ and $\nu$ are homs and hence just elements of M. $\endgroup$
    – Noah Snyder
    Commented Jun 13, 2014 at 19:10

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To give a map h(x,y,z): xyz = (xy)z -> x(yz) = xyz is to pick an element of M. This is going to give the element of H^3. The pentagon axiom becomes exactly the cocycle condition. Finally giving a tensor equivalence whose underlying functor is the identity is to give a bunch of scalars g(x,y):xy->xy and this shows that two such tensor categories are equivalent if the cocycles differ by a coboundary.

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  • $\begingroup$ Can you elaborate on how $M$ determines the map $(xy)z \mapsto x(yz)$ ? $\endgroup$
    – john mangual
    Commented Jun 15, 2014 at 0:12
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    $\begingroup$ Hom(g,g) = M for any g. In particular, Hom(xyz,xyz) = M. So to give a map from xyz to xyz is just to give an element of M. $\endgroup$
    – Noah Snyder
    Commented Jun 15, 2014 at 3:19

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