# Doubt in section 5.7 of Hoffman and Kunze's *Linear Algebra*

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 $\pi_r L= r! L$.

## 1 Answer

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.

• I agree. Hoffman & Kunze appear to be meaning this, but saying something else. – darij grinberg Jan 1 '18 at 15:23