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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):

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

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

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

  1. 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?

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    (1) What more of a difference are you looking for? (2) This seems like a language issue, but your second question is an incomplete sentence and thus not actually a question. You seem to be stating that the two equalities are analogous; are you instead asking us to confirm that they are analogous, or what? (BTW, if you think of wedges as what you get by antisymmetrizing tensors, then this Prop 3.27 follows from the analogous prop + the Leibniz formula for the determinant.) – anon Apr 28 '19 at 02:28
  • @arctictern (2) you might have missed the "is" after the "Actually," ? Anyway, I'll edit. –  Apr 28 '19 at 03:18
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    No, I didn't miss it, but I think I misinterpreted it. Now I think you're asking if the product formula is the analogue of Proposition 3.27. (I originally interpreted it as "the analogue for proposition 3.27, namely that the tensor product equals tproduct of diagonal entries, ...") I would say they're reasonably analogous, so long as you interpret the determinant with the Leibniz formula. (I can elaborate maybe tomorrow on the next comment when I don't have a headache.) – anon Apr 28 '19 at 03:27
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    But it would be more correct to say the following: tensors decompose as symmetric + antisymmetric + more components, and their corresponding effects (as multilinear functions of vectors) decompose as well. For antisymmetric tensors, the effect is by the determinant, for symmetric tensors the effect is by the permanent, and in general (the "more components" correspond to schur functors and irreps of $S_n$) the effect of a component is by an immanent. The product-of-diagonals is essentially an immanent for the regular representtion of $S_n$. – anon Apr 28 '19 at 03:31
  • @arctictern For (1), (1.1) If we have an analogue of Lemma 3.28 for tensor product, then the derivation is exactly the same except for the two differences? (1.2) Do we have an analogue of Lemma 3.28? Thanks! –  Apr 28 '19 at 03:35
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    (1.1) Yeah I believe so. (1.2) Yes for what you wrote in your question, except for the fact that $\delta_{j_1}^{i_1}\cdots\delta_{j_k}^{i_k}$ is not the same as $\delta_J^I$. For instance $\delta_1^2\delta_2^1=0$ but $\delta_{{1,2}}^{{1,2}}=1$. – anon Apr 28 '19 at 03:43
  • @arctictern Oh right. We need new notation for a compact expression. Thanks! –  Apr 28 '19 at 03:53

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