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Consider an embedding $$\Phi: G_{k_1}(R^{n_1})\times G_{k_2}(R^{n_2})\rightarrow G_k(R^n)$$ of the product of two Grassmannians $G_{k_1}(R^{n_1})\times G_{k_2}(R^{n_2})$ into $G_k(R^n)$, where $G_k(R^n)$ denotes the Grassmannian of k-plane in $R^n$.

Assume that the dimension of the product $G_{k_1}(R^{n_1})\times G_{k_2}(R^{n_2})$ plus $1$ is equal to the dimension of $G_k(R^n)$, namely $$k_1(n_1-k_1)+k_2(n_2-k_2)+1=k(n-k).$$

QUESTION: Is it true that the only possibilities for the existence of such embedding are:

either $\Phi: P^1\times P^1\rightarrow P^3$

or $\Phi: P^{n-1}\rightarrow P^n$, where $P^{n-1}=G_1(R^n)=G_{n-1}(R^n)$ denote the real projective space?

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Welcome to MO! I added the tag ag.algebraic-geometry; it is costumary to always use at least one of the top-level tags, those with a two-letter prefix corresponding to arXiv categeroies. I hope I picked the right one, you can always change this via using the link edit below the question. –  quid Jul 1 '13 at 17:24
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Are you thinking about algebraic embeddings (as suggested by quid's tags) or about smooth or topological embeddings? I suspect that all three versions are quite hard, but the algebraic version should be easier than the other two. –  Neil Strickland Jul 1 '13 at 17:29
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@Neil. The algebraic version is pretty easy: if dimension of the ambient Grassmannian is at least four, then no such embedding exists. Indeed, Picard group of any Grassmannian is $\mathbb Z$, so each divisor is ample, so Picard of each codimension one smooth subvariety is also $\mathbb Z$. –  Serge Lvovski Jul 1 '13 at 19:13
    
Thank you for your answer. I am thinking about smooth embeddings...... –  user81500 Jul 2 '13 at 9:57
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