Lemma 4.3. Aluffi Algebra Chapter VIII.

Question

The following is from the book Aluffi's “Algebra. Chapter 0” : How the (red-circled) equality holds? Esp. the l.h.s of the equality is $λ_{i_1 \dotsi_l}$ for one chosen ordered $i_1 < \dots <i_l$ but the r.h.s is a sum over all possible choices such that $i_1 < \dots <i_l$.

Answer 1

I don't like the fact that the same notation is used for the fixed index and the indexing of the summation, so let me rewrite the equality, and then explain it : the author claims that

$$\lambda_{i_1...i_l} = \overline{\varphi_I}(\sum_{1\leq j_1 < ... < j_l\leq r} \lambda_{j_1...j_l}e_{j_1}\wedge ...\wedge e_{j_l})$$

Now the reason for this is the following : $\overline{\varphi_I}$ is linear, so the rhs is $\displaystyle\sum_{1\leq j_1 < ... < j_l\leq r}\lambda_{j_1...j_l} \overline{\varphi_I}(e_{j_1}\wedge ...\wedge e_{j_l})$

But now if $(j_1...j_l)\neq (i_1...i_l)$ and they're both strictly ordered $l$-tuples, then there can be no permutation $\sigma$ such that $\sigma(j_k)=i_k$ for all $k$. Indeed such a $\sigma$ would then be an increasing bijection $\{j_1,...,j_l\}\to \{j_1,...,j_l\}$, thus (by induction on $k$ if you don't know that result) the identity, in other words $(j_1,...,j_l) = (i_1,...,i_l)$.

Therefore, by definition, for $(j_1,...,j_l)\neq (i_1,...,i_l)$, $\overline{\varphi_I}(e_{j_1}\wedge ...\wedge e_{j_l}) = \varphi_I(e_{j_1},...,e_{j_l}) = 0$.

So the only remaining index in the sum is $(i_1<...<i_l)$, and its $\overline{\varphi_I}$ is $1$ by definition, so you do get the equality

Answer 2

It's by definition of $\varphi_I$:
If $(j_1,\dots,j_\ell)=J\ne I=(i_1,\dots, i_\ell)$ and both sequences are increasing, then there is no permutation $\sigma$ of $J$ exists such that $\sigma(J)=I$, hence $$\bar\varphi_I(e_{j_1}\land\dots\land e_{j_\ell}) \ =\ 0$$