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Let $G$ be a compact real Lie group and ${\sf Trig}(G)$ the algebra of trigonometric polynomials on $G$ (defined in the Hewitt-Ross, Abstract harmonic analysis, (27.7)), i.e. the algebra of functions $u:G\to {\mathbb C}$ which can be represented as complex linear combinations of functions of the form $$ x\in G\mapsto \langle U(x)\xi,\eta\rangle\in{\mathbb C}, $$ where $U:G\to B(H)$ is an arbitrary irreducible unitary continuous representation of $G$ (in an arbitrary Hilbert space $H$), and $\xi,\eta\in H$.

Let us call a tangent vector on ${\sf Trig}(G)$ in a point $a\in G$ an arbitrary linear functional $f:{\sf Trig}(G)\to{\mathbb C}$ such that:

1) $f$ preserves involution: $$ f(\overline{u})=\overline{f(u)},\qquad u\in {\sf Trig}(G), $$

2) $f$ satisfies the Leibniz identity in the point $a$: $$ f(u\cdot v)=u(a)\cdot f(v)+f(u)\cdot v(a), \qquad u,v\in {\sf Trig}(G). $$

A question: is it true that every such functional $f:{\sf Trig}(G)\to{\mathbb C}$ is a "true tangent vector" on $G$, i.e. it can be represented in the form $$ f(u)=\lim_{t\to 0}\frac{u(\gamma(t))-u(\gamma(0))}{t},\qquad u\in {\sf Trig}(G), $$ for some smooth curve $\gamma:{\mathbb R}\to G$ with $\gamma(0)=a$?

An equivalent statement: is it true that every such functional $f:{\sf Trig}(G)\to{\mathbb C}$ can be extended to a functional $h:C^\infty(G)\to{\mathbb C}$ with the same properties?

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Can you give the definition of $Trig(G)$? – Mariano Suárez-Alvarez Jul 18 '13 at 8:39
Mariano, I have made the corrections, now ${\sf Trig}(G)$ is defined. – Sergei Akbarov Jul 18 '13 at 8:52
Usually one calls this the algebra of matrix coefficients (, yes? – Qiaochu Yuan Jul 19 '13 at 8:35
@Qiaochu Yuan: Yes, I do not see the difference. To moderators: I flagged the previous comment by chance, excuse me, that was not intentionally! – Sergei Akbarov Jul 19 '13 at 12:39
up vote 2 down vote accepted

Yes, of course. The functions you have are representation functions on $G$ (the representations are finite dimensional as well); they are thus algebraic functions on the complexification of $G$. It is well known that the "algebraic Lie algebra is the true tangent space for algebraic groups.

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Aakumadula, could you give a reference, please? – Sergei Akbarov Jul 18 '13 at 9:31
A reference for algebraic lie algebra being same as "holomorphic Lie algebra"? I cannot recall one to mind, but it is easier to give a proof. The algebraic ones give local co-ordinates; so algebraic lie algebra is $n$ dimensional, ($n=dim(G)$). Hence it has the same dimension as the "holomorphic" Lie algebra. – Venkataramana Jul 18 '13 at 13:16
Yes, this is simple... However you should have some experience in algebraic groups to feel free in this field. Can you recommend a textbook for reading and references? – Sergei Akbarov Jul 18 '13 at 16:10
I read Chevalley's book on Lie groups (see the chapter on Tannaka duality) and Humphrey's book on linear algebraic groups. Perhaps that will help? – Venkataramana Jul 19 '13 at 2:45
For me it is not obvious that ${\sf Trig}(G)$ coincides with the algebra of pokynomials on the complexification of $G$. Why is this so? – Sergei Akbarov Jul 19 '13 at 5:41

There is some difficulty. Added in edit: Only if $G$ is not compact, and infinite dimensional $H$ are also used.

  1. $G\ni x\mapsto u(x)=\langle U(x)\xi,\eta\rangle$ is in $C^\infty(G)$ if and only if one of $\xi,\eta\in H$ is a smooth vector. See section 5 of here for a complete and transparent proof.

  2. If you consider only smooth trigonometric functions, then this is true, since smooth trigonometric functions are dense in $C^\infty(G)$, and the usual proof on manifolds goes through if you replace bump functions by smooth trigonometric functions which approximate bump functions. The Taylor expansion to first order with remainder should go through in the algebra of smooth trigonometric functions.

Is this enough information, or should I try to write down a proof?

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Peter, I am not sure I understood. Are you saying that the problem appears only for non-compact $G$, or for all $G$? – Sergei Akbarov Jul 18 '13 at 9:27
Only if there are infinte dimensional irreducible unitary representations, so only for noncompact $G$. I did not notice that only had compact $G$ when I wrote my answer. – Peter Michor Jul 18 '13 at 10:00
Ah, OK! Thank you anyway! – Sergei Akbarov Jul 18 '13 at 10:04

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