I have been reading Kassel's Quantum groups and there is something I can not understand.

In Section 4 of chapter $XX$, he introduces the notion of a Infinitesimal symmetric category, that is a strict tensor category $(\mathcal{S}, \bigotimes, I)$ in which all the Hom sets are vector spaces together with, for every object $V,W$ of $\mathcal{S}$, a symmetry $\sigma_{V,W}: V \otimes W \rightarrow W \otimes V $ and an infinitesimal braiding $t_{V,W} :V \otimes W \rightarrow V \otimes W$ having the following properties:

$$\sigma_{V,W} \circ t_{V,W} = t_{W,V} \circ \sigma_{V,W}$$


$$ t_{U,V\otimes W} = t_{U,V}\otimes \mathrm{id}_W + (\sigma_{U,V}\otimes \mathrm{id}_W)^{-1} \circ(\mathrm{id}_v \otimes t_{U,W}) \circ (\sigma_{U,V}\otimes \mathrm{id}_W).$$

He then claims that if the infinitesimal symmetric category category $\mathcal{S}$ also has a left duality $V \mapsto V^*$ with structure maps $b_V^0 : I \rightarrow V \otimes V^*$ and $d_V^0 : V^* \otimes V \rightarrow I$, then we have have $$t_{V,W} = \frac12 (C_{V\otimes W}- C_V \otimes \mathrm{id}_W- \mathrm{id}_V \otimes C_W)$$ where the $(C_V : V \rightarrow V)_V$ is a natural family of endomorphisms of $\mathcal{S}$ given by : $$ C_v = - \big(\mathrm{id}_V \otimes(d_v^0 \circ t_{V^*.V})\big) \circ (b_V^0 \otimes \mathrm{id}_V).$$

I simply do not see at all why this is true, so I would glad welcome so kind of explanation. I have tried something using the fact that $(\mathrm{id}_V \otimes d_V^0) \circ (b_V^0 \otimes \mathrm{id}_V) = \mathrm{id}_V$ but this has not been very successful.

  • $\begingroup$ Did you try drawing it ? This is fairly easy to prove using graphical calculus. The following example might help your intuition: let $g$ be a Lie algebra and $t\in S^2(g)^g$ where the superscript means $g$ invariant. Then $t$ induces an infinitesimal braided structure on $g$-mod. Now write $t=\sum e_i \otimes e^i$ where $e_i$ is a basis and $e^i$ the dual basis w.r.t the pairing induced by $t$. Then C is nothing but the associated Casimir element, i.e. $C=\frac12 \sum e_ie^i$ regarded as an element in $U(g)$ (ie you replace the tensor product by the product in $U(g)$). $\endgroup$ – Adrien Apr 3 '20 at 10:33
  • $\begingroup$ Thank you for answer. I totally understand it when the category $\mathcal{S}$ is the module category for some Hopf algebra. My question is maybe to be understand in the most general setting possible, when we do not know much about $\mathcal{S}$. $\endgroup$ – Vik S. Apr 3 '20 at 11:00

Apologies for my terrible handwriting, but here is an image with a graphical proof. The first equality is one of the defining property of $t$, represented as a chord diagram, ie a dotted line between two strand (the other defining property is the obvious other version of that). For me it matter which side of a strand the dotted line is attached to, and switching from one to the other at a given vertex change the sign of the expression. Then there is the definition of $c$ (which I realize is a bit different but equivalent to yours, the end dotted line can be moved to the rightmost strand at the cost of a sign), and a computation of $c_{U\otimes V}$ which should give what you want.

Image here


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