I am trying to understand tensor products and I would like to show that $V\otimes V\simeq L(V^*,V^*,\mathbb{R})$. In Lee's book about smooth manifolds is the following proof for the case $V^*\otimes V^*\simeq L(V,V,\mathbb{R})$: enter image description here

According to the book it follows directly from $V\simeq V^{**}$, since everything is assumed to have finite dimension. I am fine with that. Nevertheless I was wondering if it couldn't be done as it is done in the above proof.

For $(v,w)\in V\times V$ and $\omega,\eta\in V^*$ I would define $\Phi :V\times V\to L(V^*,V^*,\mathbb{R}),\Phi(v,w)[\omega,\eta]:=\omega(v)\cdot\eta(w)$. I suppose it is a bilinear map. If it is, then it decents to a linear map

$\tilde{\Phi} :V\otimes V\to L(V^*,V^*,\mathbb{R}), \tilde{\Phi}(v\otimes w)[\omega,\eta]:=\omega(v)\cdot\eta(w)$. Now I am stuck. I am not sure what the basis vectors for $L(V^*,V^*,\mathbb{R})$ are or how to show that it is a bijection. I am also not sure if the apprach is okay or complete nonsense.

Thank you very much in advance!

  • 5
    $\begingroup$ Conceptually, I think it is better to deal with two different (finite-dimensional) vector spaces $V$ and $W$ to see what is really happening: show $V^* \otimes W \cong {\rm Hom}(V,W)$. In the special case $W = V$, this becomes $V^* \otimes V \cong {\rm Hom}(V,V)$. $\endgroup$
    – KCd
    Commented Feb 19, 2021 at 0:38

2 Answers 2


You just need to show that, $L(V^*,V^*,\mathbb{R})$ satisfies the universal property of tensor product.

Consider the map $\Phi :V\times V\to L(V^*,V^*,\mathbb{R}).$

Let, $W$ be a vector space with a bilinear map $B$ from $V\times V$ to $W.$

Suppose, $\{v_i^*\}$ is a basis of $V^*$ and if $v^*=\sum_{i} a^i v_i^{*},w^*=\sum_{j} b^j v_j^*,$ we have, $B'(v^*,w^*)=\sum_{i,j} a^ib^jB'(v_i^*,v_j^*)$

Then, there is a unique linear map $T :L(V^*,V^*,\mathbb{R})\to W$ defined by, $T(B')=\sum_{i,j}a^ib^jB(v_i,v_j)$ such that the diagram commutes.

Thus, from the universal property of tensor product, $V\otimes V\simeq L(V^*,V^*,\mathbb{R})$


Notation: In order to distinguish the abstract tensor product "$\otimes$" from the tensor product of covectors defined in example $12.2$, if $Z$ is a real vector space, and $f_1,f_2 \in Z^*$, let's denote by $f_1 \,\widetilde{\otimes}\, f_2$ to the bilinear map in $\mathrm L(Z,Z;\mathbb R)$ given by $(f_1 \,\widetilde{\otimes}\, f_2)(z_1,z_2) := f_1(z_1)f_2(z_2)$. Also, if $z \in Z$, let $\operatorname{ev}_z \in Z^{**}$ be "evaluation at $z$", that is, $\operatorname{ev}_z(f) = f(z)$ for every $f \in Z^*$.

Now, let $I := \{1,\dots,\dim V\}$, and suppose that $\{e_i : i \in I\}$ is a basis for $V$. Then proposition $12.8$ tells us that $\{e_\alpha \otimes e_\beta : (\alpha,\beta) \in I \times I\}$ is a basis for $V \otimes V$, and if $\{e^i : i \in I\}$ is the dual basis of $\{e_i : i \in I\}$, proposition $12.4$ tells us that $\{\operatorname{ev}_{e_\alpha} \!\widetilde{\otimes} \operatorname{ev}_{e_\beta}\! : (\alpha,\beta) \in I \times I\}$ is a basis for $\mathrm L(V^*,V^*;\mathbb R)$, simply because $\{\operatorname{ev}_{e_i}\! : i \in I\}$ is the dual basis of $\{e^i : i \in I\}$.

Finally, note that your map $\widetilde{\Phi} : V \otimes V \to \mathrm L(V^*,V^*;\mathbb R)$ sends $e_\alpha \otimes e_\beta \in V \otimes V$ to the bilinear map $\widetilde{\Phi}(e_\alpha \otimes e_\beta) \in \mathrm L(V^*,V^*;\mathbb R)$ such that $$\widetilde{\Phi}(e_\alpha \otimes e_\beta)(f_1,f_2) = f_1(e_\alpha)f_2(e_\beta) = \operatorname{ev}_{e_\alpha}(f_1) \operatorname{ev}_{e_\beta}(f_2)$$ for every $f_1$ and $f_2$ in $V^*$, in other words, $\widetilde{\Phi}(e_\alpha \otimes e_\beta) = \operatorname{ev}_{e_\alpha} \!\widetilde{\otimes} \operatorname{ev}_{e_\beta}\!$. Thus $\widetilde{\Phi}$ take basis elements in $V \otimes V$ to basis elements in $\mathrm L(V^*,V^*;\mathbb R)$, and this proves that $\widetilde{\Phi}$ is an isomorphism.

  • $\begingroup$ I think that clears up a few things.Thanks! $\endgroup$
    – Hans
    Commented Feb 19, 2021 at 14:08

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