I am wondering if it is known whether the unitary group $U(n)$ is a Kahler manifold, and, if so, what is a reference for this.
$\begingroup$
$\endgroup$
5
-
6$\begingroup$ $U(n)$ is not a complex manifold (it is defined by smooth non-holomorphic functions). More rigourously, if it were a complex manifold, then all its coordinate projections $M=(m_{ij})\mapsto m_{ij}$ would be constant by the maximum principle since $U(n)$ is compact. $\endgroup$– HenriCommented Feb 16, 2021 at 19:05
-
5$\begingroup$ For $n$ odd $U(n)$ is odd-dimensional as real manifold. By the way "is a Kähler manifold" is a quite vague question: for some given metric? diffeomorphic to a Kähler manifold? some left-invariance condition? $\endgroup$– YCorCommented Feb 16, 2021 at 19:24
-
12$\begingroup$ @Henri: $U(2)$ admits a complex structure (it is diffeomorphic to $S^1\times S^3$). $\endgroup$– Michael AlbaneseCommented Feb 16, 2021 at 19:37
-
9$\begingroup$ One could interpret the question as asking whether there is a Kähler manifold diffeomorphic to $U(n)$ for any $n$. The answer is no as $\pi_1(U(n)) \cong \mathbb{Z}$ but as is discussed here, $\mathbb{Z}$ is not a Kähler group. $\endgroup$– Michael AlbaneseCommented Feb 16, 2021 at 22:00
-
12$\begingroup$ $U(n)$ does not admit a symplectic structure because $H^2(U(n)) = 0$. $\endgroup$– user171227Commented Feb 16, 2021 at 23:53
Add a comment
|