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For a homogeneous space $M = G/H$, the number of $H$-equivariant Riemannian metrics on $M$ is usually much smaller than the space of Riemannian metrics. I am wondering what happens when the symmetric condition is relaxed, do there exist a large number of non-symmetric equivariant Riemannian metrics. Also, what is a specific example of a non-symmetric Riemannian metric on complex projective $n$-space, I am having difficulty coming up with one.

Edit: In light of the confusion my wording has caused I should make the following clarification: By {\it non-symmetric metric} I mean a collection of non-degenerate, bilinear maps for each tangent spaces $g_p:T_p(M) \times T_p(M) \to R$ for $p \in M$, inducing a map from $\chi(M) \times \chi(M) \to C^{\infty}(M)$ ($\chi(M)$ being the vector fields of $M$).

So by {\it non-symmetry} I mean that we are removing the requirement that $$ g_p(v,w) = g_p(w,v). $$

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  • $\begingroup$ What do you mean with non-symmetric? symmetric in the sense of parallel curvature tensor? $\endgroup$ – Luc Aug 7 '13 at 19:19
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I assume that, by 'equivariant metric', you mean 'invariant metric', i.e., you are looking for $G$-invariant metrics on $M=G/H$ and wondering how one describes the non-symmetric ones (when they exist). Also, I assume that you are using 'symmetric' in its standard sense, i.e., that the metric is locally symmetric, i.e., the curvature tensor of the Levi-Civita connection is parallel.

A very simple example to show that there are homogeneous metrics that are not locally symmetric is to let $G$ be a Lie group and $H$ be the identity subgroup. Then the $G$-invariant metrics on $M=G/H = G$ are exactly the left-invariant metrics, each of which is determined by knowing the metric on ${\frak{g}} = T_eG$, so there are many of these. However, unless $G$ is abelian, very few of these will be symmetric; basically, the only symmetric ones will be the ones that are also right-invariant.

More generally, the $G$-invariant metrics on $M=G/H$ are in one-to-one correspondence with the $H$-invariant positive definite inner products on the vector space $\frak{m}=\frak{g}/\frak{h}$, where the $H$-action is induced on the quotient vector space by reducing the adjoint action restricted to $H$. Thus, these correspond to an open cone (which can be nonempty if there is no $G$-invariant metric on $M$).

In the specific case of $M = \mathbb{CP}^n = \mathrm{SU}(n{+}1)/\mathrm{U}(n)$, it turns out that this cone has dimension $1$, so that, up to multiples, there is only the Fubini-Study metric, which is, of course, symmetric.

However, when $n$ is odd, one can also write $M= \mathbb{CP}^{2m-1}=\mathrm{Sp}(m)/H$ for an appropriate $H$ that acts reducibly on the quotient space $\frak{m}$. Thus, as a homogeneous space of $G = \mathrm{Sp}(m)$, the manifold $M$ admits $G$-invariant metrics that are not symmetric.

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  • $\begingroup$ Thanks you for your answer. Sorry but What I meant by symmetric was simply that, on the tangent space $T_p(M)$, at any point $p$, we $g_p(v,w) = g_p(w,v)$. Is this the same sense as your definition . . . I guess not $\endgroup$ – Noel Brown Aug 8 '13 at 11:21
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    $\begingroup$ @NoelBrown: every Riemannian metric is symmetric in your sense. Just look up the definition of Riemannian metric. $\endgroup$ – alvarezpaiva Aug 8 '13 at 11:40
  • $\begingroup$ Yes, I know, but my question was about whether there exist objects (which I decided to call "non-symetric metrics") which satisfy all the conditions of a metric except for the $g_p(v,w) = g_p(w,v)$, and which are also invariant. $\endgroup$ – Noel Brown Aug 8 '13 at 12:16
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    $\begingroup$ @Noel Brown: Your choice of terminology was unfortunate because 'non-symmetric metric' already has a well-established meaning different from what you intended. It would be better to call them something like 'bilinear form tensors' or '$(0,2)$-tensors'. In any case, for $\mathbb{CP}^n$, there are several such. For example, taking any generic, constant-coefficient linear combination of the Fubini-Study metric (which is symmetric, in both senses) and the Kähler form (which is skew-symmetric) will give you an invariant $(0,2)$-tensor that is invariant and non-symmetric in your sense. $\endgroup$ – Robert Bryant Aug 8 '13 at 12:30
  • $\begingroup$ Sorry for the confusion, and thanks a lot. $\endgroup$ – Noel Brown Aug 8 '13 at 12:41

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