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I would have a hint on how to control if the following subset $C$ is an embedded submanifold of $\mathfrak{u}(n)$.

$C$ is the set of the antihermitian $n\times n$ matrices $A$ with the property that there exists a $z\in\mathbb{C}^n\setminus{0}$ such that $A_{i,j}=\sqrt{-1}z_i^\ast z_j$ for $i,j=1,\ldots,n.$.

Thanks for any possible suggestion.


As motivation: I get this set as the image of the moment map for the natural action of $U(n)$ on $\mathbb{C}^n\setminus 0$ which is hamiltonian w.r.t. the canonical symplectic form.

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    No problem (I guessed that), I just wanted to make sure you ask what you want to ask...2011-06-13

2 Answers 2

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As Jim Belk remarks in his answer, the map $\sigma:z\in\mathbb{C}^n\to\sqrt{-1}zz^\ast\in\mathfrak{u}(n)$ is smooth and $(\mathbb{R}_+\times U(n))$-equivariant. Here the actions are given by $(t,C).z=tCz$ over $\mathbb{C}^n$, and $(t,C).A=t^2 CAC^\ast$ over $\mathfrak{u}(n)$.

Being $\{0\}$ and its complement in $\mathbb{C}^n$ the two orbits of $\mathbb{R}_+\times U(n)$, the equivariance of $\sigma$ implies that it induces a constant rank map from $\mathbb{C}^n\setminus\{0\}$ onto $\sigma(\mathbb{C}^n\setminus\{0\})$, which has to be an an orbit of $\mathbb{R}_+\times U(n)$ and so at least an initial submanifold of $\mathfrak{u}(n)$.

Being $\sigma^{-1}(\sigma(z))=\{e^{i\phi}z|\phi\in\mathbb{R}\}$ for any $z$, we get that $2n-1$ is the constant rank of $\sigma$ on $\mathbb{C}^n\setminus\{0\}$ and so even the dimension of $\sigma(\mathbb{C}^n\setminus\{0\})$ initial immersed submanifold of $\mathfrak{u}(n)$.

Finally by the observation of Jim Belk on the factorization of the action of $\mathbb{R}_+\times U(n)$ we have that $\sigma(\mathbb{C}^n\setminus\{0\}$ is an embedded submanifold of dimension $2n-1$.

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Note: The following answer is not correct, because the map $\sigma$ is not actually injective.. In particular, $\sigma(e^{i\theta}v) = \sigma(v)$ for any $v\in\mathbb{C}^n$ and $\theta \in \mathbb{R}$. See Giu's answer below.

It is easy to prove not true that the map $\sigma\colon \mathbb{C}^n \setminus 0 \to \mathfrak{u}(n)$ that you describe is an injective immersion, so the only difficult part is to show that this map is a topological embedding.

To prove this, observe that $ \mathbb{C}^n \setminus 0 \cong S^{n-1} \times (0,\infty) \qquad\text{and}\qquad \mathfrak{u}(n) \setminus 0 \cong S^d \times (0,\infty) $ where $S^{n-1}$ is the unit sphere in $\mathbb{C}^n$, and $S^d$ is the unit sphere in $\mathfrak{u}(n)$ with respect to the Euclidean norm. The map $\sigma$ carries $S^{n-1}$ to $S^d$, and satisfies $ \sigma(tu) \;=\; t^2\,\sigma(u) $ for any $u\in S^{n-1}$ and any $t>0$. Since $S^{n-1}$ is compact, the restriction of $\sigma$ to $S^{n-1}$ is an embedding, and it follows that $\sigma$ is an embedding.

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    I believe that your contribution pointed out how to proceed. Thank you. Bye!2011-06-14