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Let $X$ be a topological space (not necessarily a topological or smooth manifold). Let $f^{uv}:X\to\mathbb{R}$ be an element of $C^0=C^0(X;\mathbb{R})$, the set of all continuous real-valued functions on $X$, with $u,v\in\{1,2,\ldots,n\}$. Let $E$ be a vector space over $\mathbb{R}$ with basis $\{\mathbf{e}_1,\mathbf{e}_2,\ldots,\mathbf{e}_n\}$.

Define: \begin{gather*} \boldsymbol{\tau}:X\to\mathrm{T}^2_{\,\,0}(E)\equiv E\otimes E \quad:\Leftrightarrow\quad x\mapsto\boldsymbol{\tau}(x)=f^{uv}(x)\,\mathbf{e}_u\otimes\mathbf{e}_v\\ \boldsymbol{\tau}(x):E^*\times E^*\to\mathbb{R} \quad:\Leftrightarrow\quad (\boldsymbol{\alpha},\boldsymbol{\beta})\mapsto\boldsymbol{\tau}(x)(\boldsymbol{\alpha},\boldsymbol{\beta})=f^{uv}(x)\,\alpha_u\beta_v, \end{gather*} where $\mathrm{T}^2_{\,\,0}(E)$ denotes the set of all contravariant tensors of order 2 on $E$, $\boldsymbol{\alpha}=\alpha_u\mathbf{e}^u$, and $\boldsymbol{\beta}=\beta_v\mathbf{e}^v$. Then, is it correct to say that $\boldsymbol{\tau}$ is a $C^0$-contravariant tensor field of order 2 on $E$? That is, can $\boldsymbol{\tau}$ be a $C^0$ section of $\mathrm{T}^2_{\,\,0}(E)$? Shall I rather call it a $C^0$ section of $\mathrm{T}^2_{\,\,0}(X\times E)$, so that $\boldsymbol{\tau}$ is a $C^0$-contravariant tensor field of order 2 defined on the vector bundle $X\times E$?

In particular, if $X=\mathbb{R}^d$ and $E=\mathbb{R}^n$ with $d\neq n$, does this classification of $\boldsymbol{\tau}$ make even sense? In textbooks on differential geometry, I always find the case where the dimensions of $X$ (defined either as a topological or smooth manifold of dimension $d$) and $E=\mathbb{R}^n$ (often treated as a inner product space; a Hilbert space) are such that $d=n$, so that: \begin{equation*} \bigg\{\mathbf{e}_k\mapsto\frac{\partial}{\partial x^k}\bigg\}_{k=1}^n. \end{equation*} However, I was wondering if tensors like $\boldsymbol{\tau}$ has a special name, so that I can learn more about them.

To my understanding, $\boldsymbol{\tau}$ is a well-defined map, for given $x\in X$ and $\boldsymbol{\alpha},\boldsymbol{\beta}\in E^*$, we get indeed a real number from $\boldsymbol{\tau}(x)(\boldsymbol{\alpha},\boldsymbol{\beta})$. It is a tensor field, because $\boldsymbol{\tau}$ is a function that assigns to each $x\in X$ an element of $\mathrm{T}^2_{\,\,0}(E)$. However, what I do not know is how to classify or name this type of tensors correctly.

Thank you; Frederick.

Frederick Leuchte
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  • You have a topological space $X$ and a vector space $E$. From these, you get a trivial vector bundle $Z= X\times E$ over $X$ with fiber $E$.

  • Your map $\tau:X\to T^2_0(E)$ is technically not a section of anything. But, by a very mild reinterpretation, it can be viewed as a section of the (trivial) vector bundle $T^2_0(Z)= X\times T^2_0(E)$ over $X$, namely define $\widetilde{\tau}:X\to T^2_0(Z)$ as sending $\widetilde{\tau}(x):= (x,\tau(x))$. Barring this very common terminology and notational conflation, yes $\tau$ is (or rather gives rise to) a section of $T^2_0(Z)$.

  • You should not call $\tau$ a (2,0) tensor field on $X$ or a tensor field on $Z=X\times E$ or anything like that. In general when we speak of a $(k,l)$ tensor field on a smooth manifold $M$, we mean a (smooth) section of the vector bundle $T^k_l(TM)$ (often abbreviated to $T^k_l(M)$), the $(k,l)$-tensor bundle of $TM$

  • Suppose you have a vector bundle $(Y,\pi,B)$. A section of the vector bundle $T^k_l(Y)$ still should not be called a $(k,l)$ tensor field on $Y$, because the latter means a section of $T^k_l(TY)$ (often abbreviated to $T^k_l(Y)$… but this is obviously confusing notation when $Y$ is the total space of a vector bundle).

See Clarification on notation regarding fields, forms, and exterior algebra for more terminology and notational remarks.

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