My book is An Introduction to Manifolds by Loring W. Tu.
Here is the derivation of a basis for $A_k(V)$:
Here is the derivation of a basis for $L_k(V)$:
My first question: What's the difference besides replacing $\wedge$ with $\otimes$ and strictly ascending with arbitrary?
At first I thought that we don't have an analogue of Lemma 3.28 for $L_k(V)$ because we don't quite have an analogue of Proposition 3.27 for $L_k(V)$. But then I think we do and can prove such analogue of the lemma without any kind of analogue of the proposition (in the book, Lemma 3.28 is proved using Proposition 3.27):
$- \text{original Lemma 3.28:} \ \alpha^I(e_J) = (\alpha^{i_1} \wedge ... \wedge \alpha^{i_k})(e_{j_1}, ..., e_{j_k}) = \delta^I_J$
$- \text{analogous Lemma 3.28:} \ \alpha^I(e_J) = (\alpha^{i_1} \otimes ... \otimes \alpha^{i_k})(e_{j_1}, ..., e_{j_k}) = \alpha^{i_1}(e_{j_1}) \cdot ... \cdot \alpha^{i_k}(e_{j_k}) = \delta^{i_1}_{j_1} \cdot ... \cdot \delta^{i_k}_{j_k},$
Ligo doesn't use the notation $\alpha^I(e_J)$, but I believe Ligo's proof can be shortened with something like $\alpha^I(e_J) = \delta^I_J$.
My second question: (I intended to ask only one question, but I just thought of another) Proposition 3.27 states for wedge product of 1-covectors that wedge product equals determinant. I think the analogue for tensor product of 1-covectors is that tensor product equals product of diagonal entries as follows:
$ - \text{original Proposition 3.27 for wedge:} \ \alpha^{1} \wedge ... \wedge \alpha^{k}(v_1, ..., v_k) = \det[\alpha^{i}(v_j)]_{i,j=1,...,k}$
$ - \text{analogous Proposition 3.27 for tensor:} \ \alpha^{1} \otimes ... \otimes \alpha^{k}(v_1, ..., v_k) = \prod_{i=1,...,k} \alpha^{i}(v_i)$
Is this correct?