# Categorical description of adjoint representation

My apologies for a slightly vague question here. Let us fix a Lie algebra $$\mathfrak{g}$$ over a field $$k$$. Consider the symmetric monoidal category $$\operatorname{Rep}_{k}^{\otimes}(\mathfrak{g})$$ of representations of $$\mathfrak{g}$$. If you wish, you may consider variants of this with dg-categories, stable $$(\infty, 1)$$-categories etc. There is an element $$\mathfrak{g}^{ad}$$ of this category, moreover it is a Lie algebra object with respect to the monoidal product $$\otimes$$. Is there some way to characterize this representation categorically? Feel free to add additional structure so that the answer is positive, basically I'd like to know if there is a sensible axiomatization of "(symmetric monoidal) category with adjoint object".

• As $g^{ad}$ determines the Lie algebra structure of $g$, it does make sense to think that the category of representations is determined by $g^{ad}$. In particular: Rep(g) is enriched over itself; a g representation V is the same as a morphism of g modules $g^{ad}\to Aut_k(V)$. Can this help? Commented Jun 11, 2019 at 15:12
• @andrea thanks for this observation! I'm a bit unsure about it however. Certainly a g-rep is a hom of lie algebras g--->End(V). If we try to say this "internally" we could hope that (writing Map for the enrichment over itself) there is a unique g-module map from adj to Map(V). Is this what u're saying? I don't think it's true as can eg be seen with a one dimensional g. Nonetheless I think that since there is a natural map adj--->Map(V) maybe something can be salvaged
– user108998
Commented Jun 11, 2019 at 15:48
• If $\mathfrak{g} = \mathrm{Lie}\,G$ is the Lie algebra of some simply connected semisimple algebraic group $G$ over $k$, then one can reconstruct $G$, and thus $\mathfrak{g}$, from its monoidal category of representations using the tannakian formalism, see jmilne.org/math/xnotes/tc.pdf. I don't know if the general case can be done similarly... Commented Jun 11, 2019 at 16:13
• A more internal point of view is that $U(\mathfrak g)$ as an algebra equipped with the adjoint action can be recovered as the canonical end $\int_X End(X)$ where $End$ in general means internal endomorphism algebras, which for $\mathfrak g$-mod are just the usual $End$ equipped with the adjoint action. This should be easy to recover $g^{ad}$ from that. Commented Jun 11, 2019 at 17:00
• For $g=sl(2)$, the adjoint representation has integer spin, and so it's tensor products (and summs and summands), you miss the 1/2 spin representation $V=\mathbb C^2$, the "defining" representation of sl2. So, I doubt you could recover the complete representation theory from the adjoint representation un general. Commented Jun 13, 2019 at 20:38

Let $$(C, \otimes)$$ be an abelian complete and co-complete Cartesian-closed symmetric monoidal category over a field $$k$$. We write $$Map$$ for internal mapping objects and $$\mathbb{1}$$ for the unit.

An augmented algebra in $$(C, \otimes)$$ is an associative unital algebra object, $$X$$ together with a map of algebras $$\eta:X\rightarrow\mathbb{1}$$. We define the cotangent object of $$(X, \eta)$$ as $$\Omega_{\eta}X:= ker(\eta) /ker(\eta) ^{2}$$.

We now define the adjoint object of $$C$$ as the cotangent object of the categorical end, $$\int_{C} Map$$, at its natural augmentation. (The augmentation is induced from the map $$\int_{C} Map\rightarrow End(\mathbb{1})\cong\mathbb{1}$$).

In the case of lie algebras the end is $$(U\mathfrak{g})^{ad}$$ and so we have $$\mathfrak{g}^{ad} = \Omega_{\eta}\int_{Rep(\mathfrak{g})} Map$$.

Edit. This is incorrect, as pointed out in the comments. Adrien's comments above still suffice to produce $$U\mathfrak{g}^{ad}$$ categorically, however the attempt in this post to leverage this to a construction of $$\mathfrak{g}^{ad}$$ is flawed.

• I don't think that this is completely correct. In $U(\mathfrak{g})$, you have $ab-ba=[a,b]$, and so $[\mathfrak{g},\mathfrak{g}]\subset ker(\eta)^2$, so you will obtain the abelianisation of $\mathfrak{g}$. In general, reconstruction of $\mathfrak{g}$ from $U(\mathfrak{g})$ is a difficult problem (I believe it is partially an open question to determine if the associative algebra isomorphism $U(\mathfrak{g})\cong U(\mathfrak{h})$ implies a Lie algebra isomorphism $\mathfrak{g}\cong\mathfrak{h}$). Commented Apr 19, 2021 at 7:19
• Thank you, you are absolutely correct of course, and I've edited the post to reflect this.
– user108998
Commented Apr 19, 2021 at 9:52
• INdeed, as I said in my comment I think the trick is to consider $U_2$ the end of the $End(V\otimes W)$. Note thta this is in general larger than $U\otimes U$. Nevertheless you get by universal property a map $\Delta: U \rightarrow U_2$. In the case at hand of course this map factor through the coproduct but I don't know how to see this categorically (but you don't need that). You also have maps $l,r:U\rightarrow U_2$ from the action of $U$ on $V,W$ respectively. Then $\mathfrak g$ should be the kernel of $\Delta-l -r$. Commented Apr 19, 2021 at 10:46