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Let $G$ be a connected reductive group over $\mathbb{C}$ of (reductive) rank $\ell$. Let $P$ be a parabolic of $G$ and let $P=LN$ denote the Levi decomposition. Let $\mathfrak{g}, \mathfrak{p}, \mathfrak{l}$, and $\mathfrak{n}$ denote the corresponding Lie algebras. Recall that $N$ is normal in $P$; therefore, $P$ and $L$ act on $\mathfrak{n}$.

There is a nice description of $\mathbb{C}[\mathfrak{g}]^G$: it is a polynomial algebra on $\ell$ homogenous generators whose degrees are canonically defined.

As Steven points out, $\mathbb{C}[\mathfrak{n}]^L$ is trivial, so this means that $\mathbb{C}[\mathfrak{n}]^P$ is also trivial, since we have an inclusion $\mathbb{C}[\mathfrak{n}]^P\subset \mathbb{C}[\mathfrak{n}]^L$.

Question: Is there a description of $\mathbb{C}[\mathfrak{g}]^P$?

For instance, We have an obvious injective map $\mathbb{C}[\mathfrak{g}]^G \rightarrow \mathbb{C}[\mathfrak{g}]^P$. What is known about this map?

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  • $\begingroup$ I think you will have some trivial issues coming from tori. For example, you can pick a 1-dimensional torus $H$ in the maximal torus with respect to your data (and hence in $L$) which acts positively on all positive roots, so ${\bf C}[\mathfrak{n}]^H$ consists of scalars. $\endgroup$
    – Steven Sam
    Commented Jun 21, 2016 at 4:25

2 Answers 2

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The map $\mathbb C[\mathfrak g]^G\to\mathbb C[\mathfrak g]^P$ is an isomorphism for trivial reasons: In any quasi-affine $G$-variety, $P$ and $G$ have the same fixed points. Just look at the orbit map of a $P$-fixed point which factors through the complete variety $G/P$. Applied to representations, this means $V^G=V^P$ for any rational $G$-module. This holds for any field.

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  • $\begingroup$ Can you please elaborate? $\endgroup$
    – Vít Tuček
    Commented Jun 21, 2016 at 10:24
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    $\begingroup$ This is a nice answer. @VítTuček the answer relays on some basic facts in algebraic groups theory. 1) For every affine variety $V$ on which $G$ acts, $\mathbb{C}[V]$ is a union of $G$-invariant finite dimensional subspaces. 2) The variety $G/P$ is projective. 3) Every map from a projective variety to an affine one is constant.1+2+3) Take a $P$-fixed point in $\mathbb{C}[V]$, put it in a finite-dimensional $G$-invariant subspace, which itself is an affine variety, and consider the orbit of this point as a map from $G/P$ to it. Deduce that it is a the constant map, thus this point is $G$-fiexd. $\endgroup$
    – Uri Bader
    Commented Jun 22, 2016 at 14:41
  • $\begingroup$ Thank you @UriBader ! I wasn't aware of the third property. Unfortunately I don't have intuition for the 1 point. Do we need $G$ to be semisimple (or reductive) in order to be able to decompose $V$ into affine orbits or is it also completely general? $\endgroup$
    – Vít Tuček
    Commented Jun 22, 2016 at 15:09
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    $\begingroup$ We do not decompose $V$ into orbits, we just write the space of functions on it as a union of finite-dimensional representations. This is particularly easy if $V$ is a vector space (in your example, $V=\mathfrak{g}$): $\mathbb{C}[V]$ is the union over $d$ of the spaces of polynomials of degree no more than $d$, and each of these spaces is finite-dimensional. (Of course, you can also write it as the direct sum of the spaces of homogeneous polynomials of degree $d$.) $\endgroup$
    – t3suji
    Commented Jun 22, 2016 at 16:52
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    $\begingroup$ @t3suji: That argument works only in characteristic $0$. $\endgroup$
    – Friedrich Knop
    Commented Jun 22, 2016 at 19:41
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I also suggest reading Chapter 3 (Invariants of maximal unipotent subgroups) of:

Frank D. Grosshans, Algebraic homogeneous spaces and invariant theory, Lecture Notes in Mathematics, vol. 1673, Springer-Verlag, Berlin, 1997.

It gives an explicit description of $\mathbb{C}[\mathfrak{g}]^P$ under the adjoint action and of $\mathbb{C}[\mathfrak{g}]^N$ under left translation in the case when $\mathfrak{g}=\mathfrak{gl}_n$.

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  • $\begingroup$ Thanks Vit. Did you ever find out the answer to your question regarding $\mathbb{C}[\mathfrak{n}]^N$? $\endgroup$
    – Dr. Evil
    Commented Jun 22, 2016 at 23:23
  • $\begingroup$ @Dr.Evil No. (I assume you refer to this question: mathoverflow.net/questions/208354/…) $\endgroup$
    – Vít Tuček
    Commented Jul 13, 2016 at 15:37

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