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.
Thanks.
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 $x$.
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 $g\in G$.
