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Feb 11, 2023 at 0:35 comment added Richard Montgomery beautifully simple!
Jun 25, 2013 at 20:31 history edited Liviu Nicolaescu CC BY-SA 3.0
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Apr 18, 2012 at 4:57 history edited Liviu Nicolaescu CC BY-SA 3.0
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Apr 15, 2012 at 23:27 comment added Liviu Nicolaescu I actually discuss this in my book on Morse theory.
Apr 15, 2012 at 13:57 comment added David E Speyer And a similar trick works for the full flag manifold. Fix $\lambda_1 > \lambda_2 > \cdots > \lambda_n$ and identify $Fl_n$ with those Hermitian matrices whose eigenvalues are $\lambda_i$. Then $H \mapsto tr(A H)$ is again the desired Morse function. (The identification between Hermitian matrices and flags is as follows: Starting with a flag $F_1 \subset F_2 \subset \cdots$, let $v_i$ be an element of $F_i$ which is orthogonal to $F_{i-1}$, this is unique up to scalar. Let $H(F_\bullet})$ have eigenvector $v_i$ with eigenvalue $\lambda_i$.)
Apr 15, 2012 at 13:26 history answered Liviu Nicolaescu CC BY-SA 3.0