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Let $C$ be a curve in a projective homogeneous variety $X$.

Fixed a general point $x$ in $X$, does there exist a curve $V$ in $X$ passing through $x$ and such that $C$ and $V$ have the same homology class?

For instance when $X$ is a Grassmannian this is true.


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up vote 6 down vote accepted

I think that this problem might not be appropriate for MathOverflow, since this follows immediately from homogeneity. On the other hand, perhaps the OP is indirectly asking why every projective homogeneous variety is homogeneous under the action of a connected group. This is a standard fact in the theory of algebraic groups, but it might be worth explaining.

Let $X$ be an irreducible variety over an algebraically closed field $k$, let $G$ be a finite type group scheme over $k$, and let $\Psi:G\times_k X \to X\times_k X,$ $\Psi(g,x) = (\sigma(g,x),x)$, be an action that is surjective. Denote by $G_0 \subset G$ the identity component, and denote by $G_i$ the connected components of $G$, each of which is homogeneous under the left (or right) action of $G_0$.

The sets $\Psi(G_i \times_k X)$ are constructible subsets of $X\times_k X$ (by Chevalley) whose union equals all of $X\times_k X$ (by hypothesis). So at least one contains a dense open set. By homogeneity, every $\Psi(G_i \times_k X)$ contains a dense open set. Denote by $V_i$ the interior of $\Psi(G_i\times_k X)$, which is a dense open subset of $X\times_k X$.

Since $V_i$ and $V_0$ are dense open subsets of the irreducible variety $X\times_k X$, $V_i$ intersects $V_0$. Thus, there exists $x\in X(k)$, $g_0\in G_0(k)$, and $g_i\in G_i(k)$ such that $\sigma(g_i,x)$ equals $\sigma(g_0,x)$. So the stabilizer of $x$ intersects every connected component of $G$. Since $X$ is homogeneous, this is true for every $x\in X(k)$ (the stabilizer of $\sigma(g,x)$ is a conjugate of the stabilizer of $x$, but the property of intersecting every $G_i$ is preserved by conjugation). Thus, for every $x\in X(k)$ and for every $G_i$, there exists $g_i\in G_i(k)$ such that $\sigma(g_i,x)$ equals $x$. Since $G_i$ equals $G_0\cdot g_i$, $\sigma(G_i\times \{x\})$ equals $\sigma(G_0\times\{x\})$. Since the sets $\sigma(G_i\times\{x\})$ cover $X$ (since $X$ is homogeneous), in fact already $\sigma(G_0\times\{x\})$ equals $X$.

Now Ben Webster's argument applies. Form the family of curves in $X$, $g\cdot C$ for $g\in G_0$ parameterized by the connected variety $G_0$. For every $y\in C$, there exists $g\in G_0(k)$ such that $g\cdot y$ equals $x$. Thus $g\cdot C$ is a curve in this family that contains $x$.

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Any path connected topological group acting on a space has a trivial action on the homology, since the action of any group element is homotopic to the identity via a path in the group. Thus, if you have a homogeneous space under a path connected group, every translate of a submanifold represents the same homology class. So, you can just take $V=g\cdot C$ for an appropriate $g\in G$.

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I suspect that, secretly, the OP is asking why we can reduce the structure group to a connected group. I am just finishing an "answer" that addresses this point. – Jason Starr Mar 20 at 14:18
@JasonStarr That feels like reading in way too much thought to me. After all, the statement is obviously wrong as stated (since connectedness is not assumed). – Ben Webster Mar 20 at 17:21
Well, I am just reading it how it sounds to me. Of course I don't know what the OP means. It is true that every irreducible (finite type) scheme that is homogeneous under the action of a (finite type) group scheme is, in fact, homogeneous under the action of a connected group scheme. The first place most students realize that is when they are learning about the homology / cycle theory of projective homogeneous varieties. So I think it is a reasonable inference. – Jason Starr Mar 20 at 19:44
This is exactly the point. It was clear to me that if $X$ is homogeneous under the action of a connected group then the answer to my question was trivially yes. On the other hand I did not know that a homogeneous variety with respect to a group $G$ is always homogeneous under the action of the connected component of the identity of $G$. Thanks Jason. – JdiNirv Mar 21 at 10:03

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