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Given a variety $X$ in a projective space $\mathbb{P}^N$, $dim X=k$.

Let $A=\{(p,H): H\supset T_pX \}\subset X \times (\mathbb{P}^N)^*$.

Objective: to show that $A$ is irreducible.

Idea, way:

(1) The fibers of $\pi_1:A \longrightarrow X$ are isomorphic to $\mathbb{P}^{N-k-1}$.

(2) Item (1) implies that $A$ is irreducible.

Why is item (1) true and why does it imply that A is irreducible?

Manoel
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  • Well, for (1), what is the variety of all hyperplanes of $\mathbb{C}^{N+1}$ containing a fixed subspace $T_p$ of dimension $\mathbb{C}^{k+1}$? If you let $W$ be a subspace of $\mathbb{C}^{N+1}$ such that $\mathbb{C}^{N+1}$ is the direct sum of $T_p$ and $W$, then the dimension of $W$ is $N-k$, and the variety of all hyperplanes containing $T_p$ is in one-to-one correspondence with the variety of all hyperplanes of $W$. – Malkoun Feb 02 '20 at 17:28
  • The latter variety is of course nothing but $\mathbb{P}^{N-k-1}$. Come to think about I have made the assumption that $p$ is a smooth point of $X$, so that $T_p$ has the dimension that one would expect. – Malkoun Feb 02 '20 at 17:31
  • Negative votes sometimes seem like a lack of clarity in guiding users to better elaborate their questions. Failing to speak, a negative vote is taken. – Manoel Feb 02 '20 at 17:35
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    Forget about the negative vote. This seems like some kind of blow-up construction (or something closely related). I am not sure if this is the best way to show this, as the argument seems to assume that $X$ is a smooth variety. There may be a more direct way to show irreducibility. – Malkoun Feb 02 '20 at 17:41
  • Thanks. I'm assuming that X is non-singular, to be more precise. However, it is still true that $A={(p,H):$ p is n.s. on $X$ and $H \supset T_pX}$ is irreducible.. – Manoel Feb 02 '20 at 17:45
  • The linked question should resolve (2). If you're assuming $X$ nonsingular, then (1) is just linear algebra: you're asking "what's the dimension of the set of hyperplanes containing a $k$-dimensional subspace?" which has been addressed by Malkoun above. – KReiser Feb 02 '20 at 21:24

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