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If we have fixed a hermitian positively definite form $h(.,.)$ in complex space $C^n$ and an analytic submanifold $M$ in $C^n$, then we may fix a point outside of $M$, say $P$, and consider distance function $d(X)=h(P-X,P-X)$ for $X \in M$.

Assume that $Q$ in $M$ is a local minimum for our distance function $d$. Of course, vector $P-Q$ is orthogonal to any tangent vector $v$ to $M$ at $Q$ in euclidean sense, that is, $Re\, h(P-Q,v)=0$.

Do we actually have that $P-Q$ is orthogonal to any tangent vector $v$ at $Q$ in hermitian sense, $h(P-Q,v)=0$?

This, probably, can be seen from passing to the canonical form of $h(.,.)$, which should look like $h(u,v)=\bar{u_1}v_1+\bar{u_2}v_2+\dots+\bar{u_n}v_n$ and using the fact that $T_QM$ is a complex subspace of the vector space $C^n$, so that in appropriate coordinates $P-Q$ (the shortest direction from $P$ to $Q$) and $v$ have no nonzero coordinates in common.

My question has its origin in Milnor's "Morse theory" where he discusses Morse functions for submanifolds of euclidean or complex spaces, page 40.

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It seems that I've found a coordinate-free argument showing that $h(u,v)=0$ for $u=P-Q$ and $v\in T_QM$. Really, we need to show that $Im \, h(u,v)=0$. But $Im \, h(u,v)=Im\, h(u, (-i)iv)$, where $i^2=-1$ and so $Im \, h(u,v)=Im\, (-i)h(u,iv)=-Re\,h(u,iv)=0$, because, as I mentioned above, the tangent space $T_QM$ is a complex subspace of $C^n$ and hence is invariant under multiplication by complex numbers, which means that $iv\in T_QM$ and $Re \, h(u,iv)=0$ because $u$ and $iv$ are orthogonal in euclidean sense. –  N.B. Jan 9 '12 at 9:27

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