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I'm reading Hoffman and Kunze's Linear Algebra, 2nd edition. In $\S5.7$ "The Grassman Ring" they try to explain the way leading to the definition of exterior product (a.k.a. wedge product). However, I got a problem here that actually their calculation is different from mine in just one single step, namely for $$(D_{ij}\cdot f_k)\cdot f_l = 6\pi_4(f_i\otimes f_j\otimes f_k\otimes f_l)\tag{10}$$ my calculation shows $$(D_{ij}\cdot f_k)\cdot f_l = 12\pi_4(f_i\otimes f_j\otimes f_k\otimes f_l)\tag{10'}.$$

Can someone please throw some light on the same?

The details are given below.

First, for an alternating $r$-linear function $L\in \Lambda^r(V)$, a permutation $\sigma \in S_r$, define

$$L_\sigma (\alpha_1, \dots, \alpha_r) = L(\alpha_{\sigma 1}, \dots, \alpha_{\sigma_r})\tag{1}. $$

Then, define the $\otimes$ operation for $L \in \Lambda^r(V)$, $\alpha \in V^r$ and $M \in \Lambda^s(V)$, $\beta \in V^s$ by $$(L\otimes M)(\alpha, \beta) = L(\alpha) M(\beta). \tag{2}$$

Then, define $\pi_r$ so that $\pi_r L \in \Lambda^r(V)$ given by the formula

$$\pi_r L = \sum_\sigma (\texttt{sgn }\sigma)L_\sigma \tag{3}.$$

Then the "dot" operation $\cdot$ for $L\in \Lambda^r(V)$, $M\in \Lambda^s(V)$ (notice this $\cdot$ is not the exterior product $\wedge$ yet) given by

$$L\cdot M = \pi_{r+s}(L\otimes M) \tag{4}.$$

Notice the property $$\pi_r(L_\sigma) = (\texttt{sgn }\sigma)\pi_r(L) \tag{5}.$$

Then the following calculation defines $D_J$ for $L\in \Lambda^r(V)$: $$\begin{align} L &= \sum_J \sum_\sigma L_\sigma (\beta_{j_1}, \dots, \beta_{j_r}) f_{j_{\sigma 1}}\! \otimes \dots \otimes f_{j_{\sigma r}} \\ &=\sum_J \sum_\sigma L(\beta_{j_1}, \dots, \beta_{j_r}) (\texttt{sgn }\sigma)\ f_{j_{\sigma 1}}\! \otimes \dots \otimes f_{j_{\sigma r}} \\ &=\sum_J L(\beta_{j_1}, \dots, \beta_{j_r}) \sum_\sigma (\texttt{sgn }\sigma)\ f_{j_{\sigma 1}}\! \otimes \dots \otimes f_{j_{\sigma r}}\\ &=:\sum_J L(\beta_{j_1}, \dots, \beta_{j_r}) D_J, \tag{6} \end{align}.$$ where $J$ is the $r$-tuple $J = \{j_1, \dots, j_r\}$ such that $1\le j_1<\dots<j_r\le n$.

Some preperation for the calculation:

First $D_{ij}$, $$f_i\cdot f_j =f_i\otimes f_j - f_j \otimes f_i = D_{ij} \tag{7}.$$ Then, $D_{ij}\cdot f_k$, $$D_{ij}\cdot f_k = (f_i\cdot f_j)\cdot f_k= 2\pi_3(f_i\otimes f_j\otimes f_k) = f_i\cdot(f_j\cdot f_k) = f_i \cdot D_{jk} \tag{8}$$

Now the book is trying to conclude $D_{ij}\cdot D_{kl} \ne (D_{ij}\cdot f_k) \cdot f_l $ by showing $$D_{ij}\cdot D_{kl} = 4\pi_4(f_i\otimes f_j\otimes f_k\otimes f_l) \tag{9}$$ and $$(D_{ij}\cdot f_k)\cdot f_l = 6\pi_4(f_i\otimes f_j\otimes f_k\otimes f_l)\tag{10}.$$ I have no problem with $(9)$; however, for $(10)$ my calculation shows $$(D_{ij}\cdot f_k)\cdot f_l = 12\pi_4(f_i\otimes f_j\otimes f_k\otimes f_l)\tag{10'}.$$

Given below are the detailed steps, please help me check whether there is any error?

$$ (D_{ij}\cdot f_k)\cdot f_l \stackrel{(8)}{=} 2\pi_3(f_i\otimes f_j\otimes f_k)\cdot f_l$$

$$= 2\left(\sum_{\sigma \in S_3 =\{(1), (1,2), (1,3), (2,3), (1,2,3), (1,3,2)\}} (\texttt{sgn }\sigma) (f_i\otimes f_j\otimes f_k)_\sigma\right)\cdot f_l$$

$$= 2\left( f_i\otimes f_j\otimes f_k - f_j\otimes f_i\otimes f_k - f_k\otimes f_j\otimes f_i - f_i\otimes f_k\otimes f_j + f_k\otimes f_i\otimes f_j + f_j\otimes f_k\otimes f_i \right)\cdot f_l$$

$$\stackrel{(4)}{=} 2\left( \pi_4((f_i\otimes f_j\otimes f_k)\otimes f_l) - \pi_4((f_j\otimes f_i\otimes f_k)\otimes f_l) -\pi_4((f_k\otimes f_j\otimes f_i)\otimes f_l) - \pi_4((f_i\otimes f_k\otimes f_j)\otimes f_l) +\pi_4((f_k\otimes f_i\otimes f_j)\otimes f_l) + \pi_4((f_j\otimes f_k\otimes f_i)\otimes f_l) \right)$$ $$\stackrel{(5)}{=} 2\left( \pi_4(f_i\otimes f_j\otimes f_k\otimes f_l) + \pi_4(f_i\otimes f_j\otimes f_k\otimes f_l) + \pi_4(f_i\otimes f_j\otimes f_k\otimes f_l) + \pi_4(f_i\otimes f_j\otimes f_k\otimes f_l) + \pi_4(f_i\otimes f_j\otimes f_k\otimes f_l) + \pi_4(f_i\otimes f_j\otimes f_k\otimes f_l) \right)$$ $$=12 \pi_4(f_i\otimes f_j\otimes f_k\otimes f_l)$$

athos
  • 5,177

1 Answers1

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Your calculations are correct. This is a typo in Hoffman and Kunze's Linear Algebra. Here are a few comments regarding the calculations:

  • When listing the elements of $S_3$ under the summation symbol, you have written out the set of bijections of $\{1,2,3\}$. However, for us, the elements will be bijections of $\{ i,j,k \}$, and not $\{ 1,2,3\}$.
  • There is a slight subtlety in expanding the summation and simplfying as you have done. For example, taking $\sigma = ( i , j )$ the corresponding term is evaluated as $$\operatorname{sgn} (i,j)\ (f_i \otimes f_j \otimes f_k)_{(i,j)} = -\ f_j \otimes f_i \otimes f_k.$$ But, how exactly is this done? Are we suggesting that $$(f_i \otimes f_j \otimes f_k)_\sigma \stackrel{?}{=} (f_{\sigma i} \otimes f_{\sigma j} \otimes f_{\sigma k})$$ This is in fact not true! Instead, $$(f_i \otimes f_j \otimes f_k)_{\sigma^{-1}} = (f_{\sigma i} \otimes f_{\sigma j} \otimes f_{\sigma k}). \tag{A}$$ This identity is proved in more generality here: Proof of Theorem 7 (Chapter 5) in Hoffman and Kunze's *Linear Algebra* is unclear.

Therefore, assuming equation $(\text{A})$ to be true, we get $$(\operatorname{sgn} \sigma) (f_i \otimes f_j \otimes f_k)_\sigma = (\operatorname{sgn} \sigma) (f_{\sigma^{-1} i}\! \otimes f_{\sigma^{-1} j} \otimes f_{\sigma^{-1} k}).$$ But, since $\operatorname{sgn} \sigma = \operatorname{sgn} \sigma^{-1}$, we have $$(\operatorname{sgn} \sigma)\ (f_i \otimes f_j \otimes f_k)_\sigma = (\operatorname{sgn} \sigma^{-1})\ (f_{\sigma^{-1} i} \otimes f_{\sigma^{-1} j} \otimes f_{\sigma^{-1} k}).$$ Now, as $\sigma$ runs (once) over all permutations of $\{ i,j,k \}$, so does $\sigma^{-1}$. Therefore, $$ \sum_\sigma (\operatorname{sgn} \sigma)\ (f_i \otimes f_j \otimes f_k)_\sigma = \sum_\sigma (\operatorname{sgn} \sigma)\ f_{\sigma i} \otimes f_{\sigma j} \otimes f_{\sigma k}. $$

The above discussion also provides a cleaner proof of equation $(10\text{'})$, that is, without having to expand the sum into $6$ terms and simplify them individually. $$ \begin{align} (D_{ij} \cdot f_k) \cdot f_l &= 2\pi_3(f_i \otimes f_j \otimes f_k) \cdot f_l &&(\text{by $(8)$}) \\ &= 2 \left( \sum_{\sigma \in S_3} (\operatorname{sgn} \sigma) (f_i \otimes f_j \otimes f_k)_\sigma \right) \cdot f_l \\ &= 2 \left( \sum_{\sigma \in S_3} (\operatorname{sgn} \sigma)\ f_{\sigma i} \otimes f_{\sigma j} \otimes f_{\sigma k} \right) \cdot f_l \\ &= 2 \sum_{\sigma \in S_3} (\operatorname{sgn} \sigma) (f_{\sigma i} \otimes f_{\sigma j} \otimes f_{\sigma k}) \cdot f_l \\ &= 2 \sum_{\sigma \in S_3} (\operatorname{sgn} \sigma) \pi_4 (f_{\sigma i} \otimes f_{\sigma j} \otimes f_{\sigma k} \otimes f_l). \end{align} $$ Now, we consider $\sigma \in S_3$ to be an element of $S_4$ that fixes $l$. This clearly does not change $\operatorname{sgn} \sigma$. Thus, we have $$ \begin{align} (D_{ij} \cdot f_k) \cdot f_l &= 2 \sum_{\substack{\sigma \in S_4 \\ \sigma l = l}} (\operatorname{sgn} \sigma) \pi_4 (f_{\sigma i} \otimes f_{\sigma j} \otimes f_{\sigma k} \otimes f_{\sigma l}) \\ &= 2 \sum_{\substack{\sigma \in S_4 \\ \sigma l = l}} (\operatorname{sgn} \sigma) \pi_4 ((f_i \otimes f_j \otimes f_k \otimes f_l)_{\sigma^{-1}}) \\ &= 2 \sum_{\substack{\sigma \in S_4 \\ \sigma l = l}} (\operatorname{sgn} \sigma) (\operatorname{sgn} \sigma^{-1}) \pi_4 (f_i \otimes f_j \otimes f_k \otimes f_l) &&(\text{by $(5)$}) \\ &= 2\pi_4 (f_i \otimes f_j \otimes f_k \otimes f_l) \sum_{\sigma \in S_3} 1 \\ &= 12 \pi_4 (f_i \otimes f_j \otimes f_k \otimes f_l). \end{align} $$ Hence, proved.