# Infinitesimal categories and left duality

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:

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

and

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

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.

• 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 $U(g)$). – Adrien Apr 3 '20 at 10:33
• 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 $\mathcal{S}$. – 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.