3
$\begingroup$

The orthogonality here is determined by an alternating bilinear form $\phi$

My professor used the following definition of a "hyperbolic space" (usually this means another thing, if anyone knows how it is usually named, pls say it).

First, define "hyperbolic plan": a hyperbolic plan $H$ is a subspace of dimension two generated by two vectors not mutually orthogonal, i.e., $H=\operatorname{span}\{v,w\}$ with $\phi(v,w)\not=0$. Then a subspace $W$ is said to be hyperbolic if it is a direct sum of orthogonal hyperbolic plans,i.e., $W=H_1\oplus\cdots \oplus H_n~ \text{with}~~ H_i\bot H_j$ for each $i\not =j$.

Then, prove the following: Let $\phi$ be an alternating bilinear form, then every subspace $W$ such that $V = V^{\bot}\oplus W$ is a hyperbolic subspace.

My professor gave us the proof, by I really didn't understand anything, if anyone could explain more the steps/clarify then I'd be very thankfull!. The proof follows (I'll put some "? and ok" according to my doubts, but I didn't understand the whole thing):

"$V = V^{\bot}\oplus W \Rightarrow \phi$ not degenerated (ok). Without loss of generality suppose that $V^{\bot}=\{0\}$ ($?_1$) and we'll prove by induction on $dim(V)$.

Given $v_1 \in V\backslash \{0\}$ take $w_1$ such that $\phi(v_1,w_1)\not = -1$ (possible because $W$ not degenerated $?_2$) and $V_1=span\{v_1,w_1\}$ wich is non-degenerated $\Rightarrow V=V_1 \oplus V_1^{\bot}$ (ok)

It follows that $V_1^{\bot}$ is non-degenerated (ok). Then, by the induction hypothesis, $V_1^{\bot}$ is a hyperbolic subspace. $\blacksquare$ $(???_3)$"

I didn't get how induction was really used. Sorry for the long post, but I really need to understand this one.

$\endgroup$
1
$\begingroup$

Supposing w.l.o.g. that $V^{\perp} = \{0\}$ is a consequence of the following observation: if $V = V^{\perp}\oplus W$, then $\phi|_W$ is not degenerated. Note that we are aiming to prove that $W$ has an hyperbolic basis. By the observation above, we must deal with $\phi |_W$, which is always non-degenerated. So it's not a problem to assume that $\phi$ is indeed non-degenerated for the whole $V$.

Assumed that, it follows that exists at least one pair of vectors $v,u\in V\setminus \{0\}$ such that $\phi(v,u)=a\neq0$ , otherwise we will be contradicting the fact that $V^\perp=\{0\}$. Thefore, we can define $w=-\dfrac{1}{a}u$ and hence $\phi(v,w)=-1$, which shows that the pair $(v,w)$ is indeed hyperbolic.

Setting $V_1=[v,w]$, it follows that $V_1$ is not degenerated, because $\phi(v,w)=-1\neq0$ does ensure it. As you have notice, we can also write $V=V_1\oplus V_1^{\perp}$. The fact that $\phi$ is non-degenerate also implies that $V_1^{\perp}$ is not degenerate.

And here comes the use of induction hyphotesis. We have assumed that for all dimensions less than $n$ the claim is true. That is, for all dimensions less than $n$, if $V=W\oplus V^{\perp}$, then $W$ is hyperbolic.

Since $\mbox{dim}(V_1)=2$, we have $\mbox{dim}(V_1^{\perp})=n-2$, and this number is a suitable number for applying the induction hyphotesis. We have proven that $V_1^{\perp}$ is not degenerate. Therefore, we can write $V_1^{\perp} = W'\oplus (V_1^{\perp})^{\perp}$ for some subspace $W'$ of $V_1^{\perp}$. Now by induction hyphotesis we guarantee $W'$ hyperbolic. But since $\phi$ is not degenerate, it is true that $(V_1^{\perp})^{\perp} = V_1$. Now we write $V_1^{\perp} = W'\oplus V_1$ and this shows that $V_1^\perp$ is hyperbolic as a sum of two hyperbolic ones.

Recalling the decomposition of $V$, we have proven that $V$ is itself hyperbolic.

$\endgroup$
1
$\begingroup$

Based on your comments, I'll assume the following:

  • $V$ is a finite dimensional vector space over a field $k$ whose characteristic is not $2$.
  • $\phi:V\times V\to k$ is bilinear and $\phi(v,w)=-\phi(w,v)$
  • $V^\perp=\{v\in V\mid\phi(v,w)=0\text{ for all }w\in V\}$.

Then we can prove the statement by induction on $\dim V$. The case $V^\perp=V$ (ie $\phi=0$) is trivial since $W$ must be zero. Suppose otherwise and pick $v\notin V^\perp$. Then there exists $w\in V$ with $\phi(v,w)\neq0$. Note that $v,w$ are linearly independent since $\phi(v,v)=0$. Let $H=\mathrm{span}\{v,w\}$, which is a hyperbolic plane. Note that $\phi|_H$ is nondegenerate, so $H\cap H^\perp=0$. On the other hand $\dim H^\perp\geq\dim V-\dim H$, so $V=H\oplus H^\perp$. Thus $$ V^\perp=(H\oplus H^\perp)^\perp=H^\perp\cap(H^\perp)^\perp. $$ Applying the inductive hypothesis to $H^\perp$, we can therefore write $$ H^\perp=V^\perp\oplus W $$ where $W$ is hyperbolic. Thus $W\oplus H$ is hyperbolic and $$ V=V^\perp\oplus(W\oplus H) $$ as required.

$\endgroup$
1
$\begingroup$

Ok, the proof is indeed very rushed, so let me put a longer version of the proof. Personally, I am not a huge fan of the word "degenerated", so I will avoid that.

First, assume that the theorem holds if $V^\perp=0$. Then $\phi|_W$ is a bilinear form on $W$, and by that bilinear form, $W^\perp=0$ (easy). So by hypothesis, $W$ is hyperbolic. (this should adress $?_1$)

So now, assume $V^\perp=0$ and hence, $V=W$. Let $n=\dim(V)$. We proceed by induction to $n$. For $n=0$, there is nothing to prove. So assume $n$ is arbitrary, and assume the theorem holds for lower dimensions.

Let $v\in V\setminus\{0\}$. Then, $v\notin V^\perp$, so there exists a $w\in V$ such that $\phi(v,w)\neq 0$. Let $B$ a basis of $V$ s.t. $v\in B$. Then $V=V'\oplus H$, where $H=span(v,w)$, and:

$V'=span(\{b-\frac{\phi(b,w)}{\phi(v,w)}v-\frac{\phi(b,v)}{\phi(w,v)}w: b\in B\setminus\{v\}\})$

Then $H$ is a hyperbolic plane, $H\perp V'$, and by induction hypothesis, $V'$ is hyperbolic. So $V$ is hyperbolic. This concludes the proof.

In the rushed proof, induction is used in the same way as here, but in a hidden manner.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.