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I have a question regarding this MO answer:

The answer says that in characteristic $2$, we cannot obtain a quadratic form from a bilinear form. I thought it was the other way around and now I am confused.

I thought that if $2 \neq 0$ then we have a bijection between quadratic $q$ and bilinear forms $b$, by getting $q(x) = b(x,x)$ from $b$ and $b(x,y) = \frac{1}{2}(q(x+y) - q(x) - q(y))$.

But if $2 = 0$, we cannot divide by $2$ hence we cannot go from quadratic forms to bilinear ones, but we can still go the other way.

Would you help me resolve my confusion? Thanks a lot.

Community
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Rudy the Reindeer
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3 Answers3

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As you know, given a bilinear form $b(x,y)$, you can make a quadratic form by $q(x) = b(x,x)$. You can also make a bilinear form from a quadratic form $q(x)$ by setting $$b(x,y) = q(x+y) - q(x) - q(y).$$ Both of these operations work in arbitrary characteristic and are invariant under coordinate changes. However, they are not inverses of each other. You can easily check that a round trip of either bilinear to quadratic to bilinear or quadratic to bilinear to quadratic will introduce a factor of 2. Thus, if 2 is invertible, you can introduce a factor of one half into the definition of the bilinear form and get a bijection between quadratic and bilinear forms. On the other hand, if the characteristic is 2 then you can still convert in either direction, but both directions are neither surjective nor injective.

In the linked post, I understand the author's statement that "you can't go the other way around" not to mean that there's no way to convert a bilinear form to a quadratic form, but that there's no way to invert the map from quadratic forms to bilinear forms.

Dustin Cartwright
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Given a bilinear form $b(x,y)$, the form $x \mapsto b(x,x)$ is a quadratic form. For example, the inner product on $\mathbb{F}_2^n$ given by $( \{x_i\}, \{y_i\}) = \sum x_i y_i$ leads to the quadratic form $\sum x_i^2$, just as in characteristic $\neq 2$. But to get the bilinear form back from the quadratic form (the "norm"), you need the polarization identity, which involves factors of two.

Akhil Mathew
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  • So you're saying my understanding is correct? That would mean the answer I linked to is wrong... – Rudy the Reindeer Aug 28 '12 at 14:21
  • I cannot comment on MO, I don't have enough reputation yet (all answers I posted there are CW) – Rudy the Reindeer Aug 28 '12 at 14:23
  • @Matt: I took a look at the MO answer, and I think the answerer may have made a small error. I left a comment on it. – Akhil Mathew Aug 28 '12 at 14:44
  • Thanks a lot, Akhil! ${}{}$ – Rudy the Reindeer Aug 28 '12 at 14:46
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    @Matt No, given a quadratic $Q$ you can define a bilinear form by $B(x,y)=Q(x+y)-Q(x)-Q(y)$, and given a bilinear form you can define a quadratic form $Q(x)=B(x,x)$. Akhil is just saying that if you want these two constructions to function as inverse processes, then you need 2 to be invertible, and you use $(1/2)*(Q(x+y)-Q(x)-Q(y))=B(x,y)$ instead. – rschwieb Aug 28 '12 at 14:48
  • @rschwieb I think that's what I wrote in the OP. Except for a factor of $\frac{1}{2}$ that I think you are missing on the first line where you write $B(x,y) = Q(x+y) - Q(x) - Q(y)$. I guess I must be still missing something : ( – Rudy the Reindeer Aug 28 '12 at 14:52
  • Ok, I think Dustin's answer resolves my confusion. – Rudy the Reindeer Aug 28 '12 at 14:56
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    @matt No, nothing missing. Both of those processes are possible in any characteristic, it's just that they don't line up as inverse processes unless you scale the one in the first line by 1/2. Dustin is saying what I'm saying :) – rschwieb Aug 28 '12 at 15:03
  • @rschwieb Yes, I realised : ) Thank you! – Rudy the Reindeer Aug 28 '12 at 15:07
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    I accepted Dustin's answer now, I hope Akhil doesn't mind. Akhil's answer got 4 upvotes and Dustin's equally deserving answer 2... – Rudy the Reindeer Aug 28 '12 at 15:09
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In some areas of mathematics (group theory in particular) a quadratic form for characteristic 2 will be defined in the following way $Q:V \to k$ with

1) $Q(av)=a^2Q(v)$ for all $a \in k$ and $v \in V$ and

2) the map defined by $b(x,y)=Q(x+y)-Q(x)-Q(y)$ is bilinear

admittedly the minus signs in 2 are a little suprifulous but are meant to make it "work" in odd characteristic as well

N8tron
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  • Just to add a little bit to Nate's answer, J.H.C. Whitehead defined a quadratic morphism $Q$ between abelian groups to satisfy 1) $Q(-x)=Q(x)$, 2) $b(x,y)$ as defined above is biadditive, and then defined a universal quadratic morphism $A \to \Gamma(A)$. – Ronnie Brown Aug 28 '12 at 16:14