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Let $G$ be a compact Lie group having a left-invariant complex structure $J$.

Is there a hermitian metric $h$ in $G$, compatible with the complex structure $J$, such that $G$ is a Kähler manifold?

In the case it is not possible to turn $G$ into a Kähler manifold, does it hold that exist a hermitian metric such that $G$ is a balanced manifold? That is, $\Delta_d = 2 \Delta_\overline{\partial}$?

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To answer the first question: Only when $G$ is abelian. In fact, if $G$ is not abelian and has dimension $2d$, then there is no element of $H^2(G,\mathbb{R})$ whose $d$-th power is nonzero in $H^{2d}(G,\mathbb{R})$ (just look at the representation by bi-invariant forms), and there would have to be such an element if there were going to be even a symplectic structure on $G$, much less a Kähler structure. Conversely, if $G$ is abelian, then, yes, since $G=\mathbb{R}^{2d}/\mathbb{Z}^{2d}$, and, for any left-invariant complex structure, $G$ is $\mathbb{C}^d/\Lambda$, where $\Lambda\subset\mathbb{C}^d$ is a lattice (i.e., a discrete subgroup of maximal rank $2d$). Thus, there is a left-invariant Kähler structure, for example, (though there are others).

Added on 21 Jan 2017: As for your second question, I believe I have a partial answer, which is 'If the center of $G$ is finite (i.e., $G$ is semisimple), then $G$ does not possess a balanced metric'. Here is why: It is my understanding that, if $G$ is a compact Lie group and $J$ is a left-invariant complex structure, then $J$ is one of the complex structures constructed by Samelson and Wang. (See H.C. Wang, Closed Manifolds with Homogeneous Complex Structure, American Journal of Mathematics 76 (1954),1–32, which seems to indicate this though it doesn't state it explicitly. Also, see the article mentioned by François in the comments below.)

In such cases, there exists a maximal toral subgroup $T\subset G$ that is $J$-complex, and the quotient $G/T$ inherits a complex structure such that the quotient map $\pi:G\to G/T$ is holomorphic. Moreover, $G/T$ is not only Kähler, but an algebraic complex manifold to boot (in fact, it's the complete flag variety of $G$).

In particular, $G/T$ contains an algebraic (complex) hypersurface $Z\subset G/T$, whose preimage $X=\pi^{-1}(Z)\subset G$ is therefore a compact complex hypersurface in $G$. Now if $\omega$ were a positive $(1,1)$-form on $G$ (with respect to the complex structure $J$) such that $\omega^{d-1}$ were closed (where $2d$ is the dimension of $G$), then the integral of $\omega^{d-1}$ over $X$ would be positive. In particular, it would follow that the homology class of $X$ in $H_{2d-2}(G,\mathbb{R})$ were nonzero.

However, when $G$ is semi-simple (i.e., when its center is finite), we know by the Whitehead Lemma that $H_2(G,\mathbb{R}) = 0$, so Poincaré Duality implies that $H_{2d-2}(G,\mathbb{R})=0$ as well, contradicting the fact that the real homology class of $X$ must be nonzero.

I believe a more careful analysis of the cohomology ring of $G$ when the center of $G$ has positive dimension will lead to the result that $G$ can carry a balanced metric only when $G$ is abelian.

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    $\begingroup$ The answer to the first question goes back to Samelson (1953) (MR), who after constructing invariant complex structures on even-dimensional compact Lie groups, observes (Remark 7c): "The complex manifolds described here are in general not Kähler manifolds (for semi-simple groups the second Betti number is 0)." $\endgroup$ – Francois Ziegler Jan 15 '17 at 12:55
  • $\begingroup$ Thank you very much, Robert Bryant. This was very helpful. Regarding the second question, about balanced manifolds, should I post another question with it? It seems that it would help this site to be better organized. $\endgroup$ – Max Reinhold Jahnke Jan 15 '17 at 15:47
  • $\begingroup$ @MaxReinholJahnke: Well, you could wait to see if someone answers the balanced question, too. I haven't had time to think about it yet, but someone may have. While I'm writing, let me ask: Is the balanced condition the same as requiring that the corresponding (1,1)-form is co-closed (rather than closed)? I.e., if the manifold has dimension $2d$, we require that $\omega^{d-1}$ be closed. $\endgroup$ – Robert Bryant Jan 16 '17 at 0:15
  • $\begingroup$ Your guess that the Samelson-Wang complex structure is the only left-invariant one is right and proved in Charbonnel-Khalgui (2004) (MR) — see their remark after the theorem on p. 167. $\endgroup$ – Francois Ziegler Jan 20 '17 at 15:35
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To partial answer the first question: Theorem (A.Borel- R.Remmert) Let $X$ be a connected, compact, homogeneous Kaehler manifold. Then $X$ is isomorphic to the product $Y \times A(X)$, where $Y=G/P$ is a flag manifold and $A(X)$ is the Albanese manifold. More precisely, $Y$ is the base of the Tits fibration and the isomorphism is given by the product of the Tits fibration map and the Albanese map.

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  • $\begingroup$ This immediately gives you Robert's answer, since $Y=G/P$ can't be a compact Lie group, as the maximal compact in $P$ sits inside a maximal compact in $G$, and is not trivial as it contains the maximal compact in a Cartan subgroup. $\endgroup$ – Ben McKay Jan 15 '17 at 12:30
  • $\begingroup$ Of course that's true, but I think that the proof directly from cohomology (which only relies on the elementary Whitehead Lemma) is a bit more elementary than the Borel-Remmert Theorem. Also, the OP didn't assume that the Kähler structure was left-invariant, just that the complex structure was (although left-invariance is a bit of a red herring anyway). $\endgroup$ – Robert Bryant Jan 15 '17 at 12:44
  • $\begingroup$ Thank you very much, Ben McKay. This was helpful, but I will accept Robert Bryant's answer because I think it is easier to understand. $\endgroup$ – Max Reinhold Jahnke Jan 15 '17 at 13:21

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