Timeline for Invariant theory for parabolics
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 22, 2016 at 20:30 | comment | added | t3suji | Yes, I realize this, and it is also replacing fundamental geometric reason by something more calculation-y, but I thought it might still be worth mentioning as a possible source of intuition. | |
| Jun 22, 2016 at 19:41 | comment | added | Friedrich Knop | @t3suji: That argument works only in characteristic $0$. | |
| Jun 22, 2016 at 16:57 | comment | added | t3suji | P.S. Note that if you know the theory of representations of a reductive group, the fact that for any $G$-representation $W$, we have $W^G=W^P$ is almost obvious. Indeed, it is enough to assume that $W$ is irreducible, and then we have to show that $W^P=0$ only if $W$ is the trivial representation. But any vector in $W^P$ is clearly a highest weight vector (because it is invariant under the maximal unipotent), and its weight is zero (because it is invariant under torus), therefore, an irreducible $W$ has $W^P\ne 0$ only if $W$ is the trivial representation. | |
| Jun 22, 2016 at 16:52 | comment | added | t3suji | 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 $d$.) | |
| Jun 22, 2016 at 15:09 | comment | added | Vít Tuček | 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? | |
| Jun 22, 2016 at 14:41 | comment | added | Uri Bader | 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. | |
| Jun 21, 2016 at 10:24 | comment | added | Vít Tuček | Can you please elaborate? | |
| Jun 21, 2016 at 10:08 | vote | accept | Dr. Evil | ||
| Jun 21, 2016 at 9:12 | history | answered | Friedrich Knop | CC BY-SA 3.0 |