Codimension of Degeneracy Locus of Global Sections of a Vector Bundle

Question

Let $X$ be a variety(or scheme) of dimension $n$ and $\mathscr E$ be a vector bundle or rank $r$. Let $\tau_0,\dots,\tau_{r-i}$ for $0\leq i \leq r$ be some global sections. Then the codimension of locus of degeneracy is expected to be $i$.

My question:

1. Is there an intuitive way to see the codimension is expected to be $i$?

2. Why the following analysis is NOT correct: the locus of degeneracy is the vanishing locus of $\tau_0\wedge\dots\wedge\tau_{r-i}\in H^0(\wedge^{r-i+1}\mathscr E)$. Since $\wedge^{r-i+1}\mathscr E$ is a vector bundle of rank $\binom{r}{r-i+1}$. A global section of a rank $\binom{r}{r-i+1}$ bundle should have zero locus codimension $\binom{r}{r-i+1}$.

I know in general we can only get $codim\geq *$ result. But I'm just curious why thing goes wrong intuitively.

Answer 1

To answer the first question we first prove the following

Claim. Let $M_k(m,n)$ be the affine variety consisting of $m \times n$ matrices of rank $\leq k$. Then it has codimension $(m-k)(n-k)$ in the affine space $M=M(m,n)$ of $m\times n$ matrices.

To prove the claim, one considers the incidence correspondence $$ Y = \{(A,W) \in M \times G(n-k,n) : A \cdot W = 0\}. $$ The projection to the first factor has image equal to $M_k(m,n)$, and is one-to-one over $M_k - M_{k-1}$. Thus $\dim Y = \dim M_k(m,n)$. The variety $Y$ is smooth of dimension $k(m+n-k)$ (it is an affine space bundle over the Grassmannian).

Back to your first question: the formation of $r-i+1$ sections of $\mathcal{E}$ is equivalent to a morphism $\mathcal{O}^{r-i+1} \to \mathcal{E}$, which can be locally expressed as an $r \times (r-i+1)$ matrix with entries in $\mathcal{O}$. Thereby one obtains a map $\varphi$, locally, from $X$ into $M(r,r-i+1)$. The degeneracy locus is the preimage of $M_{r-i}(r,r-i+1)$. Thus its codimension is at least $$ (r-(r-i))(r-i+1-(r-i))= i. $$ If $\varphi$ is transverse to $M_{r-i}(r,r-i+1)$, the codimension will precisely be $i$.

For the second question: A generic section of the exterior power is not a pure tensor except when $i=0$ or $r$. Thus any (virtual or actual) expectations for "generic sections" will not be applied to special sections of the form $\tau_0 \wedge\cdots $.