Let $G(k,n)$ denote Grassmannians. For a hypersurface $W\subset \mathbb{P}^n$ of degree $2$, we let $\tau(W)\subset G(2,n+1)$ denotes the set of lines in $\mathbb{P}^n$ lying on $W$. $\tau(W)$ is an analytic cycle in $G(2,n+1)$.
It's said that since a line $l\subset \mathbb{P}^n$ lies on $W$ if and only if three points of $l$ lie on $W$, $\tau(W)$ has complex codimension $3$. I don't know why this follows? Hope someone could help. Thanks!
It's not clear to me what's the reasoning of authors, but here is my approach:
Suppose $l$ is the line joining $[1,0,...,0]$ and $[0,1,0...,0]$ on $\mathbb P^n$. Let $F(z_0,z_1,...,z_n)$ be the quadric equation of $W$, which vanishes on $l$, so there is no $z_0^2,z_0z_1,z_2^2$ terms. Moreover, up to change of coordinate, we can write $$F(z_0,...,z_n)=z_0z_2+z_1z_3+\text{higher}\ \text{order}\ \text{terms}\ \text{in}\ z_2,...,z_n.\tag{1}\label{1}$$
On the other hand, there is affine coordinate $(x_2,..,x_{n},y_2,...,y_{n})$ on $G(2,n+1)$ corresponding to the line
$$l_{x,y}=\{s[1,0,x_2,...,x_{n}]+t[0,1,y_2,...,y_{n}],\ (s,t)\in \mathbb P^1\}\tag{2}\label{2}$$ with $x_i=y_j=0$ corrsponds to $l$.
Now by plugging $(\ref{2})$ into $(\ref{1})$, the condition $l_{x,y}\subseteq W$ translates to $$F(s,t,sx_2+ty_2,...,sx_{n}+ty_{n})\equiv 0$$
Expand the polynomial and set coefficients of $s^2,st,t^2$ equal to zero, we get three equations
\begin{cases} x_2+\text{higher}\ \text{powers}=0;\\ x_3+y_2+\text{higher}\ \text{powers}=0;\\ y_3+\text{higher}\ \text{powers}=0. \end{cases}
That's the local equation of $\tau(W)$ in $G(2,n+1)$ around $l$. Since they have distinct linear terms, $\tau(W)$ is smooth of codimension three.