Tensor product and valence of a tensor?

Question

Given a vector space $V$, its dual vector space $V^{*}$ and a tensor $\mathbf{T}$:

$\mathbf{T} \in \underbrace{V \otimes\dots\otimes V}_{n\text{ copies}}\otimes \underbrace{V^{*}\otimes\dots\otimes V^{*}}_{m\text{ copies}}$

Then what is the valence $(p, q)$ of the tensor where:

• $p$ denotes the number of contravariant indices
• $q$ denotes the number of covariant indices

Is it $(n, m)$ ($n$-times contravariant and $m$-times covariant) or $(m, n)$ ($m$-times contravariant and $n$-times covariant) ?

Extra question: could someone explain me why in the wikipedia article $n$ and $m$ are swapped between the "as multilinear maps" part and the "using tensor products" part?

Answer 1

Over the reals, as an example, for a tensor $T$ to be used as a map you must consider mapping $$T: \underbrace{V^*\times...\times V^*}_{n\text{ copies}} \times \underbrace{V\times...\times V}_{m\text{ copies}}\to{\Bbb{R}}$$ via $$T(f_1,...,f_n,x_1,...,x_m)=$$ $$=T^{i_1\dots i_n}_{j_1\dots j_m}\; {e}_{i_1}\otimes\cdots\otimes{e}_{i_n}\otimes {\varepsilon}^{j_1}\otimes\cdots\otimes{\varepsilon}^{j_m}(f_1,...,f_n,x_1,...,x_m)$$ $$=T^{i_1\dots i_n}_{j_1\dots j_m}f_1(e_{i_1})\cdots f_n(e_{i_n}){\varepsilon}^{j_1}(x_1)\cdots{\varepsilon}^{j_m}(x_m),$$ where some arguments $f_i\in V^*$ and $x_j\in V$. It is routine to check multi-linearity.

Observe that evaluating at basics covectors and vectors we get: $$T({\varepsilon}^{k_1},...,{\varepsilon}^{k_n},e_{r_1},...,e_{r_m})=T^{k_1\dots k_n}_{r_1\dots r_m},$$ since ${\varepsilon}^{\alpha}(e_{\beta})={\delta^{\alpha}}_{\beta}$.

Valence and type are the same.