I am studying a proof for the dimension of $\Lambda^{\ell}(M)$ where $M$ is a rank $r$ free $R$-module from Aluffi's Algebra: Chapter 0. This is Lemma 4.3 in VIII. Linear Algebra.

The setup is that $M$ is a free module with basis $e_1,\ldots,e_r$ and the goal is to show that the generators $e_{i_1}\wedge\cdots\wedge e_{i_\ell}$ are linearly independent where $1\leqslant i_1<i_2<\cdots<i_\ell\leqslant r$. Then, for a fixed $I = (i_1,\ldots,i_\ell)$ with $1\leqslant i_1<i_2<\cdots<i_\ell\leqslant r$, he defines a function $\varphi_I:M^\ell\rightarrow R$ given by

$\varphi_I({e}_{j_1},\ldots,{e}_{j_\ell}) = \begin{cases} (-1)^\sigma & \text{if $\exists$ a permutation $\sigma$ such that $\sigma(j_k) = i_k$ for all $k$.}\\ 0 & \text{otherwise} \end{cases}$

and extending it multilinearly. He also notes that $\sigma$ is unique if it exists.

Question: Why is $\sigma$ unique?

For example, if we take $r = 7$ and consider $I = (2,5)$ and ask the value of $\varphi_I(e_2,e_3)$, then it seems as if $\sigma_1 = (3~5)$ and $\sigma_2 = (3~5)(1~6)$ both do the job of $\sigma(j_k) = i_k$ for all $k$ yet the sign of their permutations are different. What am I missing from this construction?


I think there is a typo. I am pretty sure Aluffi wants you to take $\sigma$ from $S_l$, not $S_r$.

The formula should really read

$$\varphi_I({e}_{j_1},\ldots,{e}_{j_\ell}) = \begin{cases} (-1)^\sigma & \text{if $\exists$ a permutation $\sigma\in S_l$ such that $j_{\sigma(k)} = i_k$ for all $1\leq k\leq l$.}\\ 0 & \text{otherwise} \end{cases}$$

In layman's terms, he's just asking for whether there is a rearrangement of $(j_1,j_2\cdots ,j_l)$ that yields $(i_1,\cdots i_l)$. Now it should be clear that if such a permutation exists (in $S_l$), then it is unique because we are specifying where all things should go!


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