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A quadric hypersurface (over an algebraically closed field of characteristic zero) in $\mathbb{P}^n$ for $1\leq n\leq 3$ is a toric variety. (Namely, it's isomorphic to $\mathbb{P}^1\times\mathbb{P}^1$, $\mathbb{P}^1$, or two points, depending on $n$.)

Is it true for general $n$ that quadric hypersurfaces are toric varieties? Again, over an algebraically closed field of characteristic zero.

According to Harris' book (Algebraic Geometry A First Course), on page 288, the blowup of a smooth quadric in $\mathbb{P}^n$ at a smooth point is isomorphic to the blowup of $\mathbb{P}^{n-1}$ along a smooth quadric $C$ of dimension $n-3$, but I'm not sure if that helps. And I'm assuming that the smooth case is the only interesting one, since singular quadrics are just cones over smooth ones, so I'd be content with an answer just about the smooth quadrics.

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This is false as soon as $n\geq 4$. The automorphism group of the smooth quadric in $\mathbb{P}^n$ is $PO(n+1)$; a maximal torus has dimension $[\frac{n+1}{2} ]$, which is less than $n-1$ if $n\geq 4$. – abx Feb 19 at 16:43
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Alternately, the Picard rank is $1$ as soon as $n \geq 4$ (by Lefschetz), but the only toric with picard rank $1$ is $\mathbb P^n$. – Mark Feb 19 at 16:43
    
Oh, gosh! Those are both terrific answers, and I especially should have thought of the Picard rank one myself. Thanks! – David McKinnon Feb 19 at 16:44
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(More correctly, the only smooth toric with Picard rank $1$ is $\mathbb P^n$) – Mark Feb 19 at 16:54
up vote 12 down vote accepted

(Rewriting comment for the sake of having an answer)

The quadric is not toric for $n \geq 4$. If $n \geq 4$, the Picard rank of a quadric in $\mathbb P^n$ is $1$ by the Lefschetz hyperplane theorem. The only smooth $(n-1)$-dimensional toric variety of Picard rank $1$ is $\mathbb P^{n-1}$ (see e.g. here). But a quadric in $\mathbb P^n$ isn't isomorphic to $\mathbb P^{n-1}$, for example because the canonical divisor of an $(n-1)$-dimensional quadric is divisible by $n-1$, while that of $\mathbb P^{n-1}$ isn't.

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