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Let $\mathcal{G}$ be a compact (real) Lie group. We know that the Lie algebra $\mathfrak{g}$ of $\mathcal{G}$ is, by definition, the space of all left-invariant (smooth) vector fields over $\mathcal{G}$ with bracket given by the commutator. We also know that it is isomorphic, as a Lie algebra, with $T_e\mathcal{G}$ (the tangent space to $\mathcal{G}$ at the neutral element $e$).

Consider the Hopf algebra of representative functions $H:=\mathcal{R}_{\mathbb{R}}(\mathcal{G})$ associated to $\mathcal{G}$ and recall that $H\subseteq \mathcal{C}^\infty(\mathcal{G})$ is a dense subspace with respect to the supremum norm (Proposition I.3.12 of Brocker, Dieck, Representations of Compact Lie Groups together with Peter-Weyl Throem). The Lie algebra $\mathcal{P}(H^\circ)$ of primitive elements of its finite dual is isomorphic the Lie algebra of left-invariant derivations ${^{H}\mathsf{Der}_{\mathbb{R}}(H,H)}$ (i.e., derivations $\delta:H\to H$ such that $\Delta\delta=(H\otimes \delta)\Delta$).

I would expect to have an isomorphism $\mathfrak{g}\cong {^{H}\mathsf{Der}_{\mathbb{R}}(H,H)}$.

Q1. Is this true?

I didn't find it anywhere in the literature, whence I am trying to provide one by myself.

Let $X$ be a left-invariant vector field and let $\varphi$ be a smooth function on $\mathcal{G}$. Then $\boldsymbol{X}(\varphi):\mathcal{G}\to \mathbb{R}$ given by $\boldsymbol{X}(\varphi)(g) = X_g(\varphi)$ ($X_g\in T_g\mathcal{G}$) is a smooth function and hence we have an assignment $\boldsymbol{X}:\mathcal{C}^\infty(\mathcal{G})\to \mathcal{C}^\infty(\mathcal{G})$. One can verify that this induces a Lie algebra map $$\mathfrak{g}\to {^{H}\mathsf{Der}_{\mathbb{R}}(H,H)}: X\mapsto \boldsymbol{X}.$$ To provide a candidate inverse to this morphism, I consider a left-invariant derivation $\delta$ and for every $g$ in $\mathcal{G}$ the function $\delta_g:\mathcal{R}_{\mathbb{R}}(\mathcal{G})\to \mathbb{R}, \varphi\mapsto\delta(\varphi)(g)$. Now, I would expect to be able to extend such a function to the whole $\mathcal{C}^\infty(\mathcal{G})$, maybe resorting to the Continuous Linear Extension Theorem (see Theorem 5.7.6 in Foundations of Applied Mathematics, Volume I: Mathematical Analysis by Humpherys, Jarvis, Evans), but I didn't manage to.

Q2. Could somebody suggest a way to do this?

Q3. Is there some reference in which this is treated in some detail?

OT: I already asked it on MSE, but maybe it could be that this question is more suitable for MO.

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  • $\begingroup$ As far as I know, $\mathcal{G}$ is used for Lie groupoids.. you can use normal $G$?? $\endgroup$ – Praphulla Koushik Sep 11 '18 at 10:10
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    $\begingroup$ Of course I could, but it does not make too much sense in my opinion. It's just a matter of notation and I am used to use normal $G$ for discrete groups. If you really think that using $G$ instead of $\mathcal{G}$ improves the readability of the question, you have my approval in editing it (presently it's a bit difficult for me as I don't have my laptop) $\endgroup$ – Ender Wiggins Sep 11 '18 at 11:25
  • $\begingroup$ When you define these spaces of derivations $H\to H$ are you imposing any a priori assumption of continuity? (I have faint alarm bells ringing in my head in view of the various answers and comments at mathoverflow.net/questions/6074/… but this could easily be a false alarm on my part) $\endgroup$ – Yemon Choi Jul 8 at 2:38
  • $\begingroup$ @YemonChoi No, actually I am not imposing any additional assumption apart from those explicitly mentioned. In particular, I am using Abe's book on Hopf algebras as main reference for these notions. So, derivations $\delta:H\to H$ are simply $\mathbb{R}$-linear maps such that $\delta(ab)=\delta(a)b+a\delta(b)$ (or, if you prefer, derivations from $H$ into the $H$-bimodule $H$ itself). $\endgroup$ – Ender Wiggins Jul 8 at 13:24
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Some years ago we did something that -I think- answers the algebraic version of your question [FS].

If you have a Hopf algebra $H$, then the counit gives you a map $\varepsilon_* :\mathrm{Hom}(H,H)\to \mathrm{Hom}(H,k)$. This map has a nice and natural splitting $$H^*\to \mathrm{End}(H)$$ $$f\mapsto F$$ where $F(h)=h_1f(h_2)$. This splitting actually lift to a splitting of the maps $$\varepsilon_*:\mathrm{Hom}(H^{\otimes n},H)\to \mathrm{Hom}(H^{\otimes n},k)$$ for all $n$, in a way that it is compatible with Hochschild coboundary, and stable under Gerstenhaber bracket (see the proof of Theorem 1.5 in the reference above).

In particular, $\mathrm{Der}(H,k)$ injects into $\mathrm{Der}(H,H)$. For $H=\mathcal O(G)$, $\mathrm{Der}(H,k)$ (with $k$ viewed as $H$-module via $\varepsilon$) is -almost by definition- ${\frak g}=T_e(G)$, and $\mathrm{Der}(H,H)$ is the algebraic version of ${\frak X}(G)$. The compatibility with Gerstenhaber structure says that these maps are Lie algebra maps. So you get a "Hopf"-way of defining "the unique left invariant vector field" associated to an element in the tangent space of the identity, and for $H=\mathcal O(G)$ you get a precise map $$\mathrm{Der}(\mathcal O(G),\mathcal O(G)) \cong \mathcal O(G)\otimes \mathrm{Der}(\mathcal O(G),k) =\mathcal O(G)\otimes\frak g.$$

Regarding the $C^\infty$-problems, I think that if you write the formula " $F(h)=h_1f(h_2)$" in a diagrammatic way, you should be able to do the $C^\infty$-case.

[FS] M. Farinati - A. Solotar, G-Structure on the cohomology of Hopf algebras, Proc. Amer. Math. Soc. 132 (2004), 2859-2865.] (also in https://arxiv.org/abs/math/0207243)

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  • $\begingroup$ Thank you very much, I will have a look at your reference and see if it helps. However, my main concern is that in the algebraic setting that "identification" between $T_e\mathcal{G}$, $^H\mathsf{Der}_{\Bbbk}(H,H)$ and $\mathcal{P}(H^\circ)$ is more or less well established and widely used (see Humphreys, Linear Algebraic Groups, §9.1; Hochschild, Basic Theory of Algebraic Groups and Lie Algebras, §III.3; Abe, Hopf Algebras, §4.3.1), but it seems that in the differential setting things are much more complicated (Hochschild and Mostow have a whole bunch of papers on the topic). $\endgroup$ – Ender Wiggins Jun 8 at 9:07
  • $\begingroup$ For these reasons I am really interested in the $\mathcal{C}^{\infty}$-setting $\endgroup$ – Ender Wiggins Jun 8 at 9:08

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