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 dgcategories, 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".

1$\begingroup$ 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? $\endgroup$– Andrea MarinoCommented Jun 11, 2019 at 15:12

$\begingroup$ @andrea thanks for this observation! I'm a bit unsure about it however. Certainly a grep 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 gmodule 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 $\endgroup$– user108998Commented Jun 11, 2019 at 15:48

$\begingroup$ 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... $\endgroup$– Wille LiuCommented Jun 11, 2019 at 16:13

3$\begingroup$ 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. $\endgroup$– AdrienCommented Jun 11, 2019 at 17:00

1$\begingroup$ 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. $\endgroup$– Marco FarinatiCommented Jun 13, 2019 at 20:38
1 Answer
For reference, here is the answer suggested by Adrien above.
Let $(C, \otimes) $ be an abelian complete and cocomplete Cartesianclosed 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.

2$\begingroup$ I don't think that this is completely correct. In $U(\mathfrak{g})$, you have $abba=[a,b]$, 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 $\mathfrak{g}\cong\mathfrak{h}$). $\endgroup$ Commented Apr 19, 2021 at 7:19

1$\begingroup$ Thank you, you are absolutely correct of course, and I've edited the post to reflect this. $\endgroup$– user108998Commented Apr 19, 2021 at 9:52

$\begingroup$ 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 $\Deltal r$. $\endgroup$– AdrienCommented Apr 19, 2021 at 10:46