Excuse me if this is a silly, question, but I seem to miss something obvious.

From my knowledge, given a finite-dimension vector bundle $E$ over $M$, we can always trivialize it over opens $U_{\alpha}$, and by definition; for each of these trivialisations we can find a frame $e^{\alpha}_i$. Similarly for the dual frame $E^{*}$ we find frames $e^{i}_{\alpha}$ with the property that $e^{i}_{\alpha}(x)(e^{\alpha}_{j}(x)) = \delta_{ij}$.

Now we can also form the tensor product bundle $E \otimes E^* $ which I know from references to be $\cong End(E)$.

However, to my understanding a local frame should be given by $\lbrace e^{i}_{\alpha} \otimes e^{\alpha}_{j} \;| \;1 \leq i \leq r, 1 \leq j \leq r \rbrace$, i.e.: every possible pairing of bases vectors.

It is my understanding however that we should get a frame of rank $r^2$, as that is basic linear algebra knowledge from $End(E)$, it seems however that all terms $e^{i}_{\alpha} \otimes e^{\alpha}_{j}$ where $i \neq j$ vanish, leaving us with only $r$ basis vectors.

What (I assume obvious) thing am I missing?

  • 2
    $\begingroup$ That’s actually a linear algebra question. Why do you think $e^i\otimes e_j\in E\otimes E^*$ vanishes if $i,j$ are not equal? The right argument does not act on the left, it is just their tensor product. Viewed as an endomorphism, this maps $e^j$ to $e^i\otimes e_j(e^j)=e^i\neq0$, and so it is not the zero endomorphism. $\endgroup$
    – T.P.
    Jan 2 at 20:10
  • $\begingroup$ I thought they vanish because $e^{i}_{\alpha}(x)(e^{\alpha}_{j}(x))=\delta_{ij}$. And generally speaking when one has $v \in E$ and $f \in E^{*}$ then there is the canonical evaluation map $f(v)$? But I think I understand from the comment that I should not mix up the canonical evaluation $e^{i}_{\alpha}(e^{\alpha}_j)$ with the element $e^{i} \otimes e_j$? Are my other steps then actually correct? $\endgroup$
    – Lay
    Jan 2 at 20:18
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    $\begingroup$ Yes, the rest is correct. And it is true that you should not confuse the evaluation map with the tensor. The element $v\otimes f\in E\otimes E^*$ is a formal tensor product and has nothing to do with $f(v)\in \mathbb R$ which is a simple real number (or local smooth map if we’re talking about local frames). $\endgroup$
    – T.P.
    Jan 2 at 20:23
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    $\begingroup$ I did not hide some deep meaning in my notation, I just forgot to put parentheses $(e^i\otimes e_j)(e_j)$ :D The order you write it does not really matter, it’s a matter of taste. It’s clear which element acts on what when you view it as an endomorphism. And also, as you said they are essentially the same. $\endgroup$
    – T.P.
    Jan 2 at 21:03
  • 1
    $\begingroup$ @T.P. It looks like you've essentially resolved OP's issue for them. Perhaps it'd be worthwhile for future readers to collect your comments into a full answer? $\endgroup$ Jan 3 at 3:02

1 Answer 1


I will try to clear up the confusion I personally had as much as possible in this answer: thanks to @T.P. for the insightful comments.
(I will drop $\alpha$ from the notation, may it be clear these frames are still local though. And also drop the evaluation at $(x)$, as the argument for an evaluated frame and a bases is of course the same, I will make the distinction if context is necessary.)
With these adjustments it's clear this is more of a Linear Algebra question, than it is a Differential Geometry question.

The mistake was to view $e^{i} \otimes e_{j}$ as the canonical evaluation $e^{i}(e_j)$, the latter is a smooth map when we view it as smooth frame (from which perspective the question was posed i.e.: $e^i(x)(e_j(x)) \in \mathbb{K}$, or simply as orthonormal basisvectors: $e^i(e_j)= \delta_{ij}$.)

It is important to realise $e^i \otimes e_j$ is itself an element of $E^{*} \otimes E$. However there is a canonical isomorphism to $End(E)$ following from the universal property of the tensor product. That is to say: a bilinear map $\tilde{f}: E^{*} \times E \rightarrow End(E)$ induces a unique linear map $\Phi : E^{*} \otimes E \rightarrow End(E)$ commuting with the tensor product $\otimes: (f,v) \mapsto f \otimes v$. This bilinear map is given by: $$ \tilde{f}(f,v) = f(\bullet)\cdot v\;; $$ where $ f \in E^*$ and $v \in E$, and $\bullet$ is the placeholder for an argument/element of $E$; as this should be an element in $End(E)$. So by universal property we have: $$ \Phi(f \otimes v) = f(\bullet)\cdot v .$$

Now to wrap up: the basis $e^{i} \otimes e_j$ is canonically identified with the map $e^{i}(\bullet)\cdot e_j \in End(E)$. Note it is defined on the bases vectors $e_k$ as follows: $$ e^{i}(e_k)\cdot e_j = \delta_{ik} \cdot e_j .$$ A linear map is completely determined by what it does on a basis of its codomain, hence this defines a basis of $End(E) $.
Note these are indeed $k^2$ elements: $k$ maps defining to which specific $\{e_j\}_{j= 1}^{k}$ a certain $e_i$ is sent to and then cycling through all the distinct $\{ e_i \}_{i=1}^{k}$. (For me it made it clear to concretely look at the 4 different maps we get in the case $k=2$).

To apply this to the tensor product bundle, simply replace all the bases with evaluated smooth local frames, and the transition maps with tensors of the transition maps. Also note this all generalizes to $Hom(V,W)$ instead of $End(E)$ when one chooses $V^{*}$ as $E^*$ and $W$ as $E$.


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