For simplicity, let us think of $3-$dimensional torus $\mathbb{T}^3$, which is a compact $3-$dimensional smooth manifold without boundary.

Then, we can regard the space of smooth vector fields on $\mathbb{T}^3$ simply as \begin{equation} C^\infty(\mathbb{T}^3, \mathbb{R}^3):= \{ X : \mathbb{T}^3 \to \mathbb{R}^3 \mid \text{each component of }X \text{ is a smooth real-valued function} \} \end{equation}

Now, consider $C^\infty(\mathbb{T}^3, \mathbb{R}^3)$ as a vector space over $\mathbb{R}$and let $\{ X_1, \cdots, X_n\} $ be a linearly independent subset of $C^\infty(\mathbb{T}^3, \mathbb{R}^3)$.

Then, I know that the tensor product of smooth vector fields $X,Y : M \to TM$ is defined by \begin{equation} (X \otimes Y)(p) := X_p \otimes Y_p \in T_pM \otimes T_pM \end{equation} for each $p \in M$, where $M$ is any smooth manifold.

In the case above, then the tensor products of any ordered pair from $X_1, \cdots, X_n$ live in $C^\infty(\mathbb{T}^3, \mathbb{R}^9)$. In case of $X_1$ and $X_2$, we have the smooth tensor field \begin{equation} X^i_1(x)X^j_2(x) \end{equation} where $i,j=1,2,3$ and $x \in \mathbb{T}^3$.

My question is that, are these $n^2$ tensor products linearly independent over $\mathbb{R}$ in $C^\infty(\mathbb{T}^3, \mathbb{R}^9)$?

I am not sure in the context of differential geometry, so I would like to ask to be sure..


1 Answer 1


I don't think you have to consider vector fields. Ordinary vectors in $\mathbb R^d$ are enough. For simplicity I assume $d=2\,.$ When $e_1,e_2$ is the canonical basis of $\mathbb R^2$ then the four matrices $$ e_1\otimes e_1=\begin{pmatrix}1&0\\0&0\end{pmatrix}\,,\quad e_1\otimes e_2=\begin{pmatrix}0&1\\0&0\end{pmatrix}\,,\quad e_2\otimes e_1=\begin{pmatrix}0&0\\1&0\end{pmatrix}\,,\quad e_2\otimes e_2=\begin{pmatrix}0&0\\0&1\end{pmatrix}\,,\quad $$ are obviously a basis of $\mathbb R^2\otimes\mathbb R^2\simeq\mathbb R^4\,.$

When $X_1,X_2$ are linearly independent then the matrix $$ S=\begin{pmatrix}X_{11}&X_{21}\\X_{12}&X_{22}\end{pmatrix} $$ is invertible and maps $e_i$ to $X_i\,.$

Therefore, \begin{align} 0&= \alpha_{11}\,X_1\otimes X_1+\alpha_{12}\,X_1\otimes X_2+\alpha_{21}\,X_2\otimes X_1+\alpha_{22}\,X_2\otimes X_2\\ &=S\Big(\alpha_{11}\,e_1\otimes e_1+\alpha_{12}\,e_1\otimes e_2+\alpha_{21}\,e_2\otimes e_1+\alpha_{22}\,e_2\otimes e_2)\,S^\top\,. \end{align} This implies linear independence, i.e., that $\alpha_{11}=\alpha_{12}=\alpha_{21}=\alpha_{22}=0\,.$


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