Can we talk about the $k$-planes in the product of two projective spaces?

Question

I am thinking about is there a natural way to define $k$-planes in $\mathbb{P^n}\times\mathbb{P^m}$. Here is some candidates I can come up with:

1. Being a $k$-plane is a local condition, so we can check it locally. And locally $\mathbb{P^n}\times\mathbb{P^m}$ looks the same as $(n+m)$-vector space, hence gives us a natural definition.

2. Except for some base points, there is a natural morphism $$\mathbb{P^{n+m}}\backslash B\to\mathbb{P^n}\times\mathbb{P^m}$$ so we can define a $k$-plane in $\mathbb{P^n}\times\mathbb{P^m}$ if it is the image for some $k$-plane in $\mathbb{P^{n+m}}$.

3. Using Segre embedding $$\mathbb{P^n}\times\mathbb{P^m}\to \mathbb{P^{nm+n+m}}$$ to define the $k$-planes as in $\mathbb{P^{nm+n+m}}$.

I don't know if these difinitions are equivalent. And in fact I am not satisfied with any of them. I want to know if there is a more natural way to define it?

Answer 1

Usually, to speak of a $k$-plane on an algebraic variety $X$, one needs to fix its projective embedding. If an embedding $X \subset \mathbb{P}^N$ is chosen, then a $k$-plane on $X$ is a subvariety, which when considered as a subvariety of $\mathbb{P}^N$ is a $k$-plane. This is also equivalent to its Hilbert polynomial being equal to $(t+1)(t+2)\cdots(t+k)/k!$.

In case $X = \mathbb{P}^n \times \mathbb{P}^m$, it is natural to consider the Segre embedding $\mathbb{P}^n \times \mathbb{P}^m \subset \mathbb{P}^{nm+n+m}$. Then it is easy to see that every $k$-plane lies in a fiber of either of the two natural projections of $X \to \mathbb{P}^n$ or $X \to \mathbb{P}^m$.

EDIT. The restriction of $O(1)$ from $\mathbb{P}^{nm+n+m}$ to $\mathbb{P}^k$ should be isomorphic to $O(1)$. But it restricts to $\mathbb{P}^n \times \mathbb{P}^m$ as $O(1,1)$, hence its restriction to $\mathbb{P}^k$ is the tensor product of the restriction of $O(1,0)$ and $O(0,1)$. Both these bundles are globally generated, hence so are their restrictions to $\mathbb{P}^k$. Thus, the restrictions are isomorphic to $O(k_1)$ and $O(k_2)$ with $k_1,k_2 \ge 0$, and their tensor product is $O(k_1+k_2) \cong O(1)$, hence $k_1 + k_2 = 1$. This means that either $k_1 = 0$ or $k_2 = 0$. In the first case the composition $\mathbb{P}^k \to \mathbb{P}^n \times \mathbb{P}^m \to \mathbb{P}^n$ is given by line bundle $O$, hence is constant, hence $\mathbb{P}^k$ sits in the fiber. Similarly in the second case.