I'm currently reading an article discussing flag manifolds and the action of $\mathrm{PSL}(n,\mathbb{C})$ on them. A flag (in my view at least) is a nested sequence $(y^1,\ldots,y^{n-1})$ of subspaces of $\mathbb{C}^n$ with $\dim(y^i)=i$ for all $i=1,\ldots,n-1$, and it is clear that $\mathrm{PSL}(n,\mathbb{C})$ acts transitively on these.

However, when the article discusses continuity of the action, I'm not sure what to make of it. By the orbit-stabilizer theorem, I know the space of flags (denoted by $Y$) is in one-to-one correspondence with $\mathrm{GL}(n,\mathbb{C})/\mathrm{UT}(n)$ where $\mathrm{UT}(n)$ is the group of complex upper triangular matrices. The article then claims that the action of $\mathrm{PSL}(n,\mathbb{C})$ on $Y$ is continuous with respect to both the "transcendental topology" and the Zariski topology. The transcendental topology is probably the quotient topology, induced by the above identification, but even in doing so I am not sure at all how to prove that the action is indeed continuous with respect to these topologies.

Does anybody have a hint?

  • 1
    $\begingroup$ You could also say that $Y$ is $PSL(n,\mathbb{C})/PUT(n)$ the "projective upper triangular group," since scalar matrices are upper triangular, yes? Then this is just a Lie group acting on one of its homogeneous spaces, which is certainly continuous. For the Zariski topology I think you just say $PUT$ is an algebraic subgroup of $PSL$ for which the quotient is smooth, though there are probably details here I'm missing. $\endgroup$ – Kevin Carlson May 14 '14 at 21:47
  • $\begingroup$ Wow, totally missed that. I guess the Zariski topology on $Y$ is just the quotient topology induced by the Zariski topology on $PSL(n,\mathbb{C})$ (which itself is a quotient topology). $\endgroup$ – Bryder May 14 '14 at 21:56
  • 1
    $\begingroup$ Ah yes, just for the topology that's OK. I was worries about the variety structure, which doesn't come into this claim. $\endgroup$ – Kevin Carlson May 14 '14 at 22:05
  • $\begingroup$ Ah, well, there is one concern. Two flags $x$ and $y$ are said to be transversal if $x^j\cap y^{n-j}=\{0\}$ for all $j$. For a fixed $s\in PSL(n,\mathbb{C})\setminus\{1\}$ the author then says that $\{y\in Y\,|\,sy \text{ is transversal to }y\}$ is Zariski-open in $Y$. I'm not sure if that's obvious. $\endgroup$ – Bryder May 14 '14 at 22:13
  • 1
    $\begingroup$ Well, the complement of that set is Zariski-closed: it's the union of finitely many $\{y|sy^j\cap y^{n-j}\neq \{0\}\}$. And the condition that a $j$-dimensional and an $n-j$-dimension subspace intersect nontrivially is one of the vanishing of a determinant, which is polynomial in $y^j$. $\endgroup$ – Kevin Carlson May 14 '14 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.