# Derivations of bases of $A_k(V)$ and $L_k(V)$: what's the difference?

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 $$L_k(V)$$:

1. 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):

$$- \text{original Lemma 3.28:} \ \alpha^I(e_J) = (\alpha^{i_1} \wedge ... \otimes \wedge)(e_{j_1}, ..., e_{j_k}) = \delta^I_J$$

$$- \text{analogous Lemma 3.28:} \ \alpha^I(e_J) = (\alpha^{i_1} \otimes ... \otimes \alpha^{i_k})(e_{j_1}, ..., e_{j_k}) = \alpha^{i_1}(e_{j_1}) \cdot ... \cdot \alpha^{i_k}(e_{j_k}) = \delta^{i_1}_{j_1} \cdot ... \cdot \delta^{i_k}_{j_k} = \delta^I_J,$$

• Ligo doesn't use the notation $$\alpha^I(e_J)$$, but I believe Ligo's proof can be shortened with something like $$\alpha^I(e_J) = \delta^I_J$$.

1. My second question: (I intended to ask only one question, but I just thought of another) Actually, is the analogue for Proposition 3.27 that tensor product equals product of diagonal entries, which is analogous to wedge product equals determinant in the original Proposition 3.27?

$$- \text{original Proposition 3.27:} \ \alpha^{1} \wedge ... \wedge \alpha^{k}(v_1, ..., v_k) = \det[\alpha^{i}(v_j)]_{i,j=1,...,k}$$

$$- \text{analogous Proposition 3.27:} \ \alpha^{1} \otimes ... \otimes \alpha^{k}(v_1, ..., v_k) = \prod_{i=1,...,k} \alpha^{i}(v_i)$$