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In section 5.7 (The Grassman Ring) of Hoffman and Kunze's Linear Algebra, the authors write (on page 174)

The proof of the lemma following equation $(5\text{-}36)$ shows that for any $r$-linear form $L$ and any permutation $\sigma$ of $\{1,\dots,r\}$ $$\pi_r(L_\sigma) = \operatorname{sgn}{\sigma}\ \pi_r(L)$$

It is not clear to me how the given equation follows from the proof of the lemma following equation $(5\text{-}36)$. Can someone explain the above statement to me?

The relevant lemma and notations are given below.


Notations and definitions

Let $V$ be a free module of rank $n$ over a commutative ring $K$ with identity. We denote the space of all $r$-linear forms on $V$ by $M^r(V)$ and the space of all alternating $r$-linear forms by $\Lambda^r(V)$. For $L \in M^r(V)$ and any permutation $\sigma$ of $\{1,\dots,r\}$, we obtain another $r$-linear function $L_\sigma$ by defining $$L_\sigma(\alpha_1,\dots,\alpha_r) = L(\alpha_{\sigma 1},\dots,\alpha_{\sigma r})$$ for all $(\alpha_1,\dots,\alpha_r) \in V^r$. For each $L \in M^r(V)$, we define the alternating $r$-linear function $\pi_r L$ by $$\pi_r L = \sum_\sigma (\operatorname{sgn} \sigma) L_\sigma$$ where the sum is over all permutations $\sigma$ of $\{1,\dots,r\}$.


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. Let $\tau$ be any permutation of $\{1,\dots,r\}$. Then $$ \begin{align} (\pi_r L)(\alpha_{\tau 1},\dots,\alpha_{\tau r}) &= \sum_\sigma (\operatorname{sgn} \sigma)\ L(\alpha_{\tau \sigma 1}, \dots, \alpha_{\tau \sigma r}) \\ &= (\operatorname{sgn} \tau) \sum_\sigma (\operatorname{sgn} \tau\sigma)\ L(\alpha_{\tau \sigma 1},\dots,\alpha_{\tau \sigma r}). \end{align} $$ As $\sigma$ runs (once) over all permutations of $\{1,\dots,r\}$, so does $\tau\sigma$. Therefore, $$ (\pi_r L)(\alpha_{\tau 1},\dots,\alpha_{\tau r}) = (\operatorname{sgn} \tau)(\pi_r L)(\alpha_1,\dots,\alpha_r). $$ Thus, $\pi_r L$ is an alternating form.

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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2 Answers 2

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The proof of the lemma shows that for $L \in M^r(V)$ and $\tau$ a permutation of $\{1,\dots,r\}$, we have $$(\pi_r L)_\tau = (\operatorname{sgn}{\tau}) (\pi_r L).$$ This precisely says that $\pi_r L \in \Lambda^r(V)$, which is the statement of the lemma. (As an aside, note that the authors are implicitly assuming that $K$ is a ring in which $1+1 \neq 0$; for more details, see Why is $\pi_r(L)$ a linear transformation into $\Lambda^r(V)$.)

This is not the same as $\pi_r(L_\sigma) = \operatorname{sgn}{\sigma}\ \pi_r L$, so this does not follow from the proof of the lemma.


To prove the given identity, let $(\alpha_1,\dots,\alpha_r) \in V^r$. Then, $$ \begin{align} \pi_r (L_\sigma)(\alpha_1,\dots,\alpha_r) &= \sum_\tau (\operatorname{sgn}{\tau})L_\sigma(\alpha_{\tau 1},\dots,\alpha_{\tau r})\\ &= \sum_{\tau} (\operatorname{sgn}{\tau})L(\alpha_{\tau\sigma 1},\dots,\alpha_{\tau\sigma r})\\ &= (\operatorname{sgn}{\sigma})\sum_{\tau}(\operatorname{sgn}{\tau\sigma})L(\alpha_{\tau\sigma 1},\dots,\alpha_{\tau\sigma r}). \end{align} $$ As $\tau$ runs (once) over all the permutations of $\{ 1,\dots,r \}$, so does $\tau\sigma$. Therefore, $$ \pi_r(L_\sigma)(\alpha_1,\dots,\alpha_r) = (\operatorname{sgn} \sigma)(\pi_r L)(\alpha_1,\dots,\alpha_r). $$ Since $(\alpha_1,\dots,\alpha_r)$ was an arbitrary element of $V^r$, we have $$ \pi_r (L_\sigma) = \operatorname{sgn}{\sigma}\ \pi_r(L). $$


One can see that the idea of this proof is the same as in the proof of the lemma. Perhaps that was what the authors meant in their statement.

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    $\begingroup$ I agree. Hoffman & Kunze appear to be meaning this, but saying something else. $\endgroup$ Jan 1, 2018 at 15:23
  • $\begingroup$ I would say that what the proof of the lemma shows is actually showing $\pi_r (L_\tau) = (sgn \tau) (\pi_r(L))$ since we have $\pi_r (L_\tau) (\alpha_1 ,..., \alpha_r) = (\sum_{\sigma} (sgn \sigma) (L_\tau)_\sigma) (\alpha_1 ,..., \alpha_r) $ and although stated implicitly here, the authors stated this also again explicitly in the next subchapter that $(N_\tau)_\sigma = N_{\sigma \tau}$ so that $\endgroup$
    – hteica
    Apr 6, 2020 at 12:22
  • $\begingroup$ so that $\pi_r (L_\tau) (\alpha_1 ,..., \alpha_r) = (\sum_{\sigma} (sgn \sigma) (L_{\tau})_\sigma) (\alpha_1 ,..., \alpha_r) = (\sum_{\sigma} (sgn \sigma) (L_{\sigma \tau})) (\alpha_1 ,..., \alpha_r) = (\sum_{\sigma} (sgn \sigma) (L_{\sigma})) (\alpha_{\tau 1} ,..., \alpha_ {\tau r}) = (\pi_r L) (\alpha_{\tau 1} ,..., \alpha_ {\tau r}) $ . Note that according to the author, $(\pi_r L)_\tau (\alpha_1,...,\alpha_r) \neq (\pi_r L) (\alpha_{\tau 1},...,\alpha_{\tau r})$, I admit that it's confusing at first, but it becomes clear after the explanation in the next chapter. $\endgroup$
    – hteica
    Apr 6, 2020 at 12:32
  • $\begingroup$ @hteica I agree, it's implicit in the proof. I say in my answer that the idea of the proof is the same as in the lemma, too. But it would improve the exposition if this were stated more explicitly here itself. (On the other hand, we have Math SE for ironing out these details. :)) $\endgroup$
    – user279515
    Apr 6, 2020 at 12:54
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    $\begingroup$ Ok done, hopefully it helps. $\endgroup$
    – hteica
    Apr 6, 2020 at 14:07
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The proof of the lemma shows that $$\pi_r(L_\tau) = \operatorname{sgn}{\tau}\ \pi_r L$$ Why? We first note that $(N_{\tau})_{\sigma} = N_{\sigma \tau}$ (stated explicitly in the subchapter)

So we have:

\begin{align} \pi_r (L_\tau)(\alpha_1,\dots,\alpha_r) &= (\sum_\sigma (\operatorname{sgn}{\sigma})(L_\tau)_\sigma)(\alpha_{1},\dots,\alpha_{r})\\ &= (\sum_\sigma (\operatorname{sgn}{\sigma})(L_{\sigma \tau}))(\alpha_{1},\dots,\alpha_{r})\\ &= (\sum_\sigma (\operatorname{sgn}{\sigma})(L_{\sigma}))(\alpha_{ \tau 1},\dots,\alpha_{ \tau r})\\ &= \pi_r (L)(\alpha_{\tau 1},\dots,\alpha_{\tau r}) . \end{align}

And the rest follows exactly as in the proof of the lemma (stated also in question above).

Maybe an important point to note :

$(\pi_r L)_\tau (\alpha_1,...,\alpha_r) \neq (\pi_r L) (\alpha_{\tau 1},...,\alpha_{\tau r})$.

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