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Let $k$ be an algebraically closed field of characteristic $\neq 2$ and $Q_r\subset \mathbb P^n_k$ the quadric hypersurface given by the equation $x_0^2+\cdots+x_r^2=0 \;(0\leq r\leq n)$ in the projective $n$-space over $k$ with coordinates $x_0, \cdots,x_n$.
Needless to say $Q_r$ is singular if $r\lt n$.

Question: what is the maximum dimension $M=M(n,r)$ of the linear subspaces $L\subset Q_r$, as a function of $n$ and $r$?
For example $M(3,3)=1$, $M(2,1)=1$ and $M(n,0)=n-1$.

lefuneste
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  • $\lfloor (r+1)/2 \rfloor + (n - r) - 1$. – Sasha Aug 17 '19 at 11:59
  • @Sasha: This is most probably correct because it is compatible with the numerous special cases I checked. Could you please give a reference or sketch a proof in an answer, so that I can accept it and close the discussion? Anyway, congratulations and many thanks for this quick and precise answer. – lefuneste Aug 17 '19 at 14:03
  • @Sasha: Дорогой Саша, вы русский? Во всяком случае я уже давно заметил, что вы отличный математик. – lefuneste Aug 18 '19 at 11:52

1 Answers1

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First note, that for a nondegenerate quadratic form on a vector space $V$ the maximal dimension of an isotropic subspace is $$ \lfloor \dim(V)/2 \rfloor. $$ Indeed, if there is a bigger isotropic subspace, it is easy to see that it contains a kernel vector of the form. On the other hand, the existence of such isotropic subspace can be easily shown by induction on $\dim(V)$.

Next, if a quadratic form is degenerate and $K \subset V$ is its kernel space, then a maximal isotropic subspace $I \subset V$ contains $K$ (otherwise $I + K$ is a bigger isotropic subspace).

Finally, isotropic subspaces in $V$ containing $K$ are in bijection with isotropic subspaces in $V/K$ (for the induced quadratic form) via the map, that takes $I \subset V$ to $I/K \subset V/K$.

Combining all this, it is clear that the dimension of a maximal subspace is $$ \dim K + \lfloor \dim(V/K)/2 \rfloor. $$ Subtracting 1 to get the dimension of the associated projective space, one obtains $$ (n-r) + \lfloor (r+1)/2 \rfloor - 1. $$

Sasha
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  • Please don't confuse the quadratic form with the nondegenerate bilinear form $b(u,v) = \sum_{j=0}^r u_jv_j$ on $k^{r+1}$. If we have $b(W,W)=0,\dim(W) > (r+1)/2$ how do you get a contradiction ? – reuns Aug 17 '19 at 16:05
  • @reuns: Since the characteristic is not 2, there is no difference between quadratic and bilinear forms. If $b(W,W) = 0$, then the map $V \to V^\vee$ induced by $B$ takes $W \subset V$ to $W^\perp \subset V^\vee$. If $\dim(W) > (r+1)/2$ then $\dim(W) > \dim(W^\perp)$, hence the induced map $W \to W^\perp$ has a nonzero kernel, which is a kernel space for $b$. – Sasha Aug 17 '19 at 19:45
  • Thanks a lot for your wonderfully clear and definitive answer Sasha. – lefuneste Aug 17 '19 at 20:14
  • Dear Sasha, I have answered an arbitrary question on the site in order to have enough reputation for being allowed to upvote you :-). – lefuneste Aug 18 '19 at 08:58