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.
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6$U(n)$ 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. – Henri Feb 16 '21 at 19:05
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5For $n$ 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? – YCor Feb 16 '21 at 19:24
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12@Henri: $U(2)$ admits a complex structure (it is diffeomorphic to $S^1\times S^3$). – Michael Albanese Feb 16 '21 at 19:37
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9One 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. – Michael Albanese Feb 16 '21 at 22:00
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12$U(n)$ does not admit a symplectic structure because $H^2(U(n)) = 0$. – user171227 Feb 16 '21 at 23:53