Product Basis?

**Showcase_22** · Dec 19th 2009, 05:12 AM

I have this theorem:

Theorem 33.2: Let $\{e_i \}$ and $\{ e^i \}$ be dual bases for $V$ and $V^*$ . Then the set of tensor products:

$\{ e_{i_1} \otimes \ldots \otimes e_{i_p} \otimes e^{j_1} \otimes \ldots \otimes e^{j_q}, \ i_1, \ldots , i_p,j_1, \ldots , j_q=1, \ldots , N \}$

forms a basis for $\vartheta_q^p(V)$ (the set of all tensors of order $(p,q)$ on a vector space $V$ ).

The proof starts by proving that the set is linearly independent. It does this as follows:

"We shall prove that the set of tensor products is a linearly independent generating set for $\vartheta_q^p(V)$ . To prove that the set is linearly independent, let

$A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q} e_{i_1} \otimes \ldots \otimes e_{i_p} \otimes e^{j_1} \otimes \ldots \otimes e^{j_q}=0$

where the RHS is the 0 tensor".

Firstly, what is $A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q} e_{i_1}$ ? Is it a linear map going from $V$ to $V^*$ with the bases outlined in the question?

Secondly, how will this prove that the set is linearly independent?

**Opalg** · Dec 19th 2009, 07:45 AM

Originally Posted by Showcase_22

I have this theorem:

The proof starts by proving that the set is linearly independent. It does this as follows:

"We shall prove that the set of tensor products is a linearly independent generating set for $\vartheta_q^p(V)$ . To prove that the set is linearly independent, let

$A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q} e_{i_1} \otimes \ldots \otimes e_{i_p} \otimes e^{j_1} \otimes \ldots \otimes e^{j_q}=0$

where the RHS is the 0 tensor".

Firstly, what is $A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q} e_{i_1}$ ? Is it a linear map going from $V$ to $V^*$ with the bases outlined in the question?

Secondly, how will this prove that the set is linearly independent?

$A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q}$ is a scalar, for each set of indices $\{i_1, \ldots, i_p,j_1, \ldots , j_q\}$ . You should read this expression as $\sum A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q}(e_{i_1} \otimes \ldots \otimes e_{i_p} \otimes e^{j_1} \otimes \ldots \otimes e^{j_q})$ . The summation convention is being used, which means that you are meant to sum (from 1 to N) over each index that appears both as a superscript and as a subscript. So that expression simply denotes a linear combination of the given tensors. To prove that they are linearly independent, you put a linear combination of them equal to 0, and then you are aiming to show that each coefficient $A^{i_1, \ldots, i_p}_{j_1, \ldots , j_q}$ is zero.

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