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I'm reading Linear Algebra by Hoffman and Kunze, and there's a proof that just doesn't seem correct.

He wants to prove that $\pi_r(L)$ a linear transformation from $M^r(V)$ into $\Lambda^r(V)$. In the proof, he establishes that if $\tau$ is a permutation of $\{1,2,\ldots,r\}$ then $\pi_rL(\alpha_{\tau_1},\ldots,\alpha_{\tau_r})=(sgn\ \tau)L(\alpha_1,\ldots,\alpha_r)$. He then concludes that $\pi_rL$ is an alternating form

Now, if $L$ is an alternating form then $L_\alpha=(sgn\ \alpha)L$. However, the converse isn't necessarily true, is it? Can someone please explain to me what I didn't understand?

It's on page 170, if that would be of any help.

Thanks in advance.

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    $\begingroup$ Can you detail the context a bit more? Notably, does the discussion allow for characteristic$~2$, or is that excluded? In characteristic$~2$ being alternating requires more than "changing sign" for odd permutations of arguments, but in other characteristics the two are equivalent. $\endgroup$ – Marc van Leeuwen Feb 9 '15 at 13:00
  • $\begingroup$ That is the problem. He didn't explicitly exclude fields of characteristic 2. I was thinking that maybe something holds in this case even in characteristic 2. So, do I understand that this proof is wrong in fields of characteristic two as I thought, right? That's all the context there was. (Except that it was towards any commutative ring with identity, which can be a field of characteristic 2 anyway.) $\endgroup$ – Hasan Saad Feb 9 '15 at 13:08
  • $\begingroup$ @MarcvanLeeuwen You might want to put what you're going to say as an answer so we can mark this question as solved. :) $\endgroup$ – Hasan Saad Feb 9 '15 at 13:14
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    $\begingroup$ This is a bit surprising, since Hoffman and Kunze are usually very careful about characteristic. In fact, on p.144, the authors have stated explicitly that $1+1=0$, the definition using swapping is weaker to the definition using duplicates. No idea why they trip over here. Anyway, although the proof is incorrect/incomplete, the lemma itself is correct. $\endgroup$ – user1551 Jan 2 '18 at 10:33
  • $\begingroup$ @user1551 I am decidedly against having a tag specific to a single book. Please stop adding that tag unless the meta discussion concludes that it is a good idea. My inclination is to just remove that tag from every post, but the community decides, and you are welcome to weigh in in meta. $\endgroup$ – Jyrki Lahtonen Jan 2 '18 at 12:33
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To conclude that a multilinear form is alternating because permuting the arguments by a permutation$~\sigma$ multiplies its values by $\operatorname{sgn} \sigma$ is not valid in contexts where the scalar $2$ is not necessarily invertible (for instance when discussing vector spaces over a field that might be of characteristic$~2$). Basically in characteristic$~2$ that condition means being symmetric, which is (there) strictly weaker than being alternating, with an example showing the difference already arising for bilinear forms in dimension$~1$ (where an alternating bilinear form is necessarily$~0$, but a symmetric one is not).

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For the sake of completeness, I would like to add a proof of this lemma that does not assume that $K$ is a ring in which $1 + 1 \neq 0$.

Lemma. $\pi_r$ is a linear transformation from $M^r(V)$ into $\Lambda^r(V)$. If $L$ is in $\Lambda^r(V)$ then $\pi_r L = r! L$.

Proof. It is clear that $\pi_r$ is a linear transformation from $M^r(V)$ into $M^r(V)$. Let $(\alpha_1,\dots,\alpha_r) \in V^r$ and suppose that $\alpha_i = \alpha_j$ for some $1 \leq i < j \leq r$. We will show that $\pi_r L(\alpha_1,\dots,\alpha_r) = 0$, proving that $\pi_r L$ is alternating.

We pair the permutations of $\{ 1,\dots,r \}$ in a manner such that if $\sigma$ and $\tau$ are paired, then $$(\operatorname{sgn}{\sigma})L(\alpha_{\sigma 1},\dots,\alpha_{\sigma r}) + (\operatorname{sgn}{\tau})L(\alpha_{\tau 1},\dots,\alpha_{\tau r}) = 0.$$ The pairing is as follows: pair $\sigma$ with $\tau = (i,j) \circ \sigma$. This is well-defined because by this rule $\tau$ is paired with $$(i,j) \circ \tau = (i,j) \circ (i,j) \circ \sigma = \sigma,$$ and $\sigma \neq \tau$ because $\operatorname{sgn}{\sigma} =-\operatorname{sgn}{\tau}$.

If $\sigma^{-1} i < \sigma^{-1} j$, we get $$ \begin{align} (\operatorname{sgn}{\tau})L(\alpha_{\tau 1},\dots,\alpha_{\tau r}) &= -(\operatorname{sgn}{\sigma})L(\alpha_{(i,j) \circ \sigma 1}, \dots, \alpha_{(i,j) i}, \dots, \alpha_{(i,j) j}, \dots, \alpha_{(i,j) \circ \sigma r})\\ &= -(\operatorname{sgn}{\sigma}) L(\alpha_{\sigma 1}, \dots, \alpha_{j}, \dots, \alpha_{i}, \dots, \alpha_{\sigma r}) \\ &= -(\operatorname{sgn}{\sigma}) L(\alpha_{\sigma 1}, \dots, \alpha_{i}, \dots, \alpha_{j}, \dots, \alpha_{\sigma r}) &&(\because \alpha_i = \alpha_j)\\ &= -(\operatorname{sgn}{\sigma})L(\alpha_{\sigma 1},\dots,\alpha_{\sigma r}) \end{align} $$ If $\sigma^{-1} j < \sigma^{-1} i$, we get $$ \begin{align} (\operatorname{sgn}{\tau})L(\alpha_{\tau 1},\dots,\alpha_{\tau r}) &= -(\operatorname{sgn}{\sigma})L(\alpha_{(i,j) \circ \sigma 1}, \dots, \alpha_{(i,j) j}, \dots, \alpha_{(i,j) i}, \dots, \alpha_{(i,j) \circ \sigma r})\\ &= -(\operatorname{sgn}{\sigma}) L(\alpha_{\sigma 1}, \dots, \alpha_{i}, \dots, \alpha_{j}, \dots, \alpha_{\sigma r}) \\ &= -(\operatorname{sgn}{\sigma}) L(\alpha_{\sigma 1}, \dots, \alpha_{j}, \dots, \alpha_{i}, \dots, \alpha_{\sigma r}) &&(\because \alpha_i = \alpha_j)\\ &= -(\operatorname{sgn}{\sigma})L(\alpha_{\sigma 1},\dots,\alpha_{\sigma r}) \end{align} $$ In either case, $$(\operatorname{sgn}{\sigma})L(\alpha_{\sigma 1},\dots,\alpha_{\sigma r}) + (\operatorname{sgn}{\tau})L(\alpha_{\tau 1},\dots,\alpha_{\tau r}) = 0.$$ Therefore, $$ \pi_r L(\alpha_1,\dots,\alpha_r) = \sum_{\sigma} (\operatorname{sgn}{\sigma})L(\alpha_{\sigma 1},\dots,\alpha_{\sigma r}) = 0. $$ So, $\pi_r$ is a linear transformation from $M^r(V)$ into $\Lambda^r(V)$.

If $L$ is in $\Lambda^r(V)$, then $L(\alpha_{\sigma 1},\dots,\alpha_{\sigma r}) = (\operatorname{sgn} \sigma) L(\alpha_1,\dots,\alpha_r)$ for each $\sigma$; hence $\pi_r L= r! L$.

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