I encountered a problem which confuses me, that is, it is whether or not Tensor product of two vector fields turns out a vector field?

For example:

Let $M$ be a smoth manifold, $C^{\infty}\left(M\right)$ denote the commutative ring of smoth functions on $M$ and $C^{\infty}\left(TM\right)$ be the set of smoth vector fields on $M$ forming a module over $C^{\infty}\left(M\right).$ Put $C_0^{\infty}\left(TM\right)=C^{\infty}\left(M\right)$ and for each positive integer $r$ let $$C_r^{\infty}\left(TM\right)= C^{\infty}\left(TM\right) \otimes \ldots \otimes C^{\infty}\left(TM\right)$$ be $r$-fold tensor product of $C^{\infty}\left(TM\right)$ over $C^{\infty}\left(M\right).$


Let $M$ be a differentiable manifold. A smoth tensor field $T$ on $M$ of type $\left(r,s\right)$ is a map $T:C_r^{\infty}\left(TM\right) \rightarrow C_s^{\infty}\left(TM\right)$ which is multi-linear over $C^{\infty}\left(M\right)$ i.e. $$T\left( X_1 \otimes \dots X_{k-1} \otimes \left( f.Y+g.Z \right) \otimes X_{k+1} \otimes \dots \otimes X_r \right) $$ $$=f.T\left( X_1 \otimes \dots X_{k-1} \otimes Y \otimes X_{k+1} \otimes \dots \otimes X_r \right)+g.T\left( X_1 \otimes \dots X_{k-1} \otimes Z \otimes X_{k+1} \otimes \dots \otimes X_r \right)$$ for all $X_1, \dots, X_r,Y,Z \in C^{\infty}\left(TM\right)$ and $f,g \in C^{\infty}\left(M\right).$

It follows from the above deffinition that whether or not $T\left( X_1 , \dots , X_r \right)$ is a vector field? Thank you!

  • 1
    $\begingroup$ No. It is a tesnor product of $s$ vector fields. If $s=1$, then yes. $\endgroup$ – user98602 Dec 28 '16 at 19:37

Let us study the simplest case: consider $X, Y \in C^\infty (TM)$; you are asking whether $X \otimes Y \in C^\infty (TM)$.

  • What is $X \otimes Y$? It is a smooth map that takes $x \in M$ into $X_x \otimes Y_x \in T_x M \otimes T_x M$.

  • What is a vector field $Z$? It is a smooth map that takes $x \in M$ into $Z_x \in T_x M$.

Since $T_x M \otimes T_x M \ne T_x M$ (unless $\dim M = 1$), we cannot call elements of $T_x M \otimes T_x M$ tangent vectors at $x$, therefore we cannot call elements of $C^\infty (TM) \otimes C^\infty (TM)$ tangent fields (or 'vector fields", these two are synonymous). We call them tensor fields.

Similarly, if $T$ is a $(r,s)$-tensor field, then $T(X_1, \dots, X_r)$ is a $s$-tensor field.

Notice, though, that if $T$ is a $(0,1)$-tensor field, it may be naturally viewed as a vector field, but this is the only such case.

  • $\begingroup$ yeah! thank you so much! $\endgroup$ – Hao Tran Dec 28 '16 at 20:08

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