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I am reading the book algebraic geometry: a first course by Joe Harris. I have a question about codimension of a variety.

Usually, if we have a subspace $W$ of dim $k$ in a space $V$ of dim $n$, then $W$ is of codimension $n-k$. But I cannot compute the codimension on line -4 above Exercise 11.43 on page 149 of the book. It is said that codim of $\Psi$ in $\mathbb{G}(k,n)$ is $(k+1)*(k-(m+l-n))$. I think that $dim(\mathbb{G}(k,n))=(k+1)*(n-k)$. Since it is shown that $dim(\Psi)=(k+1)(m-k)+(l-k)(n-l)$, codimension of $\Psi$ in $\mathbb{G}(k,n)$ should be $(k+1)*(n-k)-((k+1)(m-k)+(l-k)(n-l))$. But this is not $(k+1)*(k-(m+l-n))$.

Thank you very much.

Edit: here $\Psi=\{(\Gamma, \Theta): \Theta \subset \Gamma\} \subset \mathbb{G}(k,\Lambda) \times \mathbb{G}(l,n)$. $\Lambda \subset \mathbb{P}^n$ is a fixed $m$-plane. We need to compute the dimension of $\Sigma_{k}(\Lambda)=\{\Gamma: dim(\Gamma \cap \Lambda) \geq k\} \subset \mathbb{G}(l,n)$. Why introduce $\Psi$ to compute the dimension? Here $\mathbb{G}(k,n)$ is the set of all $k$-plane in $\mathbb{P}^n$.

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Please don't use Google Books links; they are very unstable, and what one can see often depends on where one is located geographically, so that many users of this website will not be able to see what you are seeing in that link. It's better to give the bibliographic citation, and quote the problem. – Arturo Magidin Sep 10 '11 at 22:45
@Arturo, thanks. How to use bibliographic citation? – LJR Sep 11 '11 at 0:06
By "bibliographic citation" I mean: give the name of the book, the author, and the page number/exercise number/theorem number (and the edition, if necessary). That way anyone with access to the book can see what you are seeing. – Arturo Magidin Sep 11 '11 at 0:07
up vote 0 down vote accepted

It seems that there are some typos. The statement should be the codimension of $\Sigma_k$ in $\mathbb{G}(l,n)$ is $(k+1)(k-(m+l-n))$ and $\Psi$ should be $\{(\Theta, \Gamma): \Theta \subset \Gamma\} \subset \mathbb{G}(k, \Lambda) \times \mathbb{G}(l,n)$.

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