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?


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.