Timeline for Enveloping algebras of map algebras as hyperalgebras of algebraic groups
Current License: CC BY-SA 3.0
13 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Dec 26, 2013 at 22:36 | comment | added | Chuck Hague | Ok, thanks - that's very helpful! I appreciate it. I'll check out the reference and think about how to lift the Lie algebra statement to a hyperalgebraic statement. | |
| Dec 26, 2013 at 21:14 | comment | added | user76758 | By the way, for a literature reference on the Lie algebra of ${\rm{R}}_{A/k}(G_A)$ for affine algebraic $k$-groups $G$ (or even ${\rm{R}}_{A/k}(\mathcal{G})$ for affine algebraic $A$-groups $\mathcal{G}$), see the more-or-less self-contained section A.7 (esp. Cor. A.7.6) in the book "Pseudo-reductive groups". | |
| Dec 26, 2013 at 21:10 | comment | added | user76758 | Oops, that was silly of me to overlook! Here is an idea over fields $k$ of characteristic 0. Let $H := {\rm{Spec}}(U(\mathfrak{g}_A))$ for a $k$-finite $A$. We seek an isomorphism of affine algebraic $k$-groups $H\simeq{\rm{R}}_{A/k}(G_A)$ extending the "identity" on Lie algebras. This is unique if it exists (since the target is Zariski-connected). Using an inclusion of algebraic groups $G\hookrightarrow{\rm{SL}}_n$, perhaps we can reduce to the case of ${\rm{SL}}_n$ via Lie algebra considerations (at the cost of directly proving $U(\mathfrak{g}_A)$ is a domain with "expected dimension"). | |
| Dec 26, 2013 at 19:16 | comment | added | Chuck Hague | I think $U(\mathfrak g_A)$ sees more than just the underlying vector space of $A$; multiplication in $A$ is part of the Lie bracket in $\mathfrak g_A$, so commutation in $U$ sees the algebra structure of $A$. For example, for the dual numbers $A = \mathbb C[x] / \langle x^2 \rangle$, the bracket relations in $\mathfrak g_A$ give a natural isomorphism $ \mathfrak g_A \cong \mathfrak g \ltimes \mathfrak g^{ab} $, where by $\mathfrak g^{ab}$ I mean the abelian Lie algebra with underlying space $\mathfrak g$. And this is indeed the Lie algebra of the tangent bundle $G \ltimes \mathfrak g$ of $G$. | |
| Dec 26, 2013 at 18:51 | comment | added | user76758 | For any field $k$, $k$-finite $A$, and affine algebraic $k$-group scheme $G$, ${\rm{Lie}}({\rm{R}}_{A/k}(G_A))$ is the underlying Lie algebra over $k$ of $\mathfrak{g}_A$. But if we regard $\mathfrak{g}_A$ as a Lie algebra over $k$ then $U(\mathfrak{g}_A)$ only "knows" $A$ through its underlying $k$-vector space, so it is $U(\mathfrak{g}^{\oplus n}) = U(\mathfrak{g})^{\otimes n}$ (with $n=\dim_k A$) and thus recovers $G^n$ in your setup. In contrast, the group structure of ${\rm{R}}_{A/k}(G_A)(B)=G(A\otimes_k B)$ involves the $k$-algebra structure of $A$; compare $A=k\times k, k[x]/(x^2)$. | |
| Dec 26, 2013 at 18:09 | comment | added | Chuck Hague | Actually, your comment about the Weil restriction for $\mathbb C$-finite $A$ is part of the motivation for my question - as a particular special case I've been trying to navigate the literature on jet group schemes (so this is the case where $A = \mathbb C[t]/ \langle t^n \rangle$ for some $n > 0$ and then considering the Weil restriction). It seems implicit in some of the literature that we do indeed obtain this enveloping algebra $U$ in this case - for example, it's true when we consider the dual numbers - but I'm having a hard time navigating the literature. | |
| Dec 26, 2013 at 18:00 | comment | added | user76758 | Oh, OK, so you're taking the Lie algebra $\mathfrak{g}_A$ over $A$ and viewing it instead just as a Lie algebra over $\mathbf{C}$ for the formation of the universal enveloping algebra (so it isn't an $A$-algebra, etc.). Then your question makes sense, though for $\mathbf{C}$-finite $A$ do you know if the Weil restriction ${\rm{R}}_{A/\mathbf{C}}(G_A)$ has coordinate ring equal to that enveloping algebra? (I would be surprised if this is true, but maybe there is a simple trick.) | |
| Dec 26, 2013 at 17:40 | comment | added | Chuck Hague | I'm considering $\mathfrak g_A$ purely as a Lie algebra (in general infinite-dimensional) over $\mathbb C$, in which case I would expect that the augmentation $\epsilon$ should take values in $\mathbb C$, not $A$, unless I'm confused here? The element $aX$ doesn't make sense in $U$ as far as I can tell; that is, I don't expect $U$ to be an $A$-algebra. Only $X \otimes a$ makes sense, for $X \in \mathfrak g$ and $a \in A$, in which case $\Delta(X \otimes a) = (X \otimes a) \otimes 1 + 1 \otimes (X \otimes a)$. Thanks for the EGA reference - I'll look at that. | |
| Dec 26, 2013 at 16:41 | comment | added | user76758 | Dear Chuck: Your $\epsilon$ defines a homomorphism from $U:=U(\mathfrak{g}_A)$ into $A$, not into $\mathbf{C}$; what does $\epsilon$ do to the natural image of $A$ in $U$? Likewise, your $\Delta$ is a map from the $A$-algebra $U$ into $U\otimes_A U$, not into $U \otimes_{\mathbf{C}}U$; if you think $\Delta$ is a map into the latter then where do $aX$ and $a$ go under this map for $a \in A$? For representability when $A=\mathbf{C}[t]$, if $G=\mathbf{G}_a$ then this amounts to a "universal polynomial" in $t$; considering nilpotent coefficients rules it out (EGA IV$_3$, 8.14.2 makes it rigorous). | |
| Dec 26, 2013 at 16:07 | comment | added | Chuck Hague | I'm thinking of the Hopf algebra structure where we put $\epsilon(X) = 0$ and $\Delta X = X \otimes 1 + 1 \otimes X$ for all $X \in \mathfrak g_A \subseteq U(\mathfrak g_A)$ and then extend multiplicatively. Thanks for the info on representability - is there a good reference for reading about representability/non-representability of functors like this? | |
| Dec 26, 2013 at 12:37 | comment | added | user76758 | A (co-commutatie) Hopf algebra $R$ over $\mathbf{C}$ is equipped with an "identity section" $R \rightarrow \mathbf{C}$ as $\mathbf{C}$-algebras and a co-multiplication $R \rightarrow R \otimes_{\mathbf{C}} R$ as $\mathbf{C}$-algebras. For the Hopf algebra $U(\mathfrak{g}_A)$ over $A$, what such data do you have in mind? Also, your proposed functor fails to be representable for $A=\mathbf{C}[t]$ and $G$ any linear algebraic $\mathbf{C}$-group with positive dimension. (Weil restriction through an algebra map not of finite dimension arises for "affine Grassmannians", which are not representable.) | |
| Dec 25, 2013 at 22:24 | history | asked | Chuck Hague | CC BY-SA 3.0 |