# Tensor product commutes with pullback?

• CompuChip
In summary, the conversation discusses an exercise involving pulling back a metric under a function and the assumption that the pullback commutes with the tensor product. The speaker manages to complete the exercise using this assumption and clarifies the steps involved, including the use of definitions for the tensor product and pullback. They also mention their lack of familiarity with pulls and pushes and ask for further clarification.
CompuChip
Homework Helper
Hello,

I have an exercise where we have to pullback a metric $g^{ij} \, \mathrm dx_i \, \mathrm dx_j$ under a function $f: M \rightarrow N$ (actually in this case $M = \mathbf{R}^2, N = \mathbf{R}^3$, but that's not really relevant).

I managed to do it, provided that the pullback commutes with the tensor product. That is, using that actually $\mathrm dx_i \, \mathrm dx_j = \mathrm dx_i \otimes \mathrm dx_j$, I assumed that
$$f^*(\mathrm dx_i \, \mathrm dx_j) = (f^*(\mathrm dx_i)) \otimes (f^*(\mathrm dx_j))$$
so then I could use that
$$f^*(\mathrm dx_i) = \mathrm d(f^* x_i) = \mathrm df_i = \frac{\partial f_i}{\partial x_k} \mathrm dx_k$$
and finish the exercise.

Why is this true?

Thanks a lot.

$$f^*(dx_i\otimes dx_j)(u,v)=dx_i\otimes dx_j (f_*u,f_*v)=dx_i(f_*u) dx_j(f_*v)$$

$$=f^*(dx_i)(u)f^*(dx_j)(v)=f^*(dx_i)\otimes f^*(dx_j)(u,v)$$

Thank you very much!

I missed a lot on this topic though, so I'm not really familiar with pulls/pushes. Could you clarify the first (= third) step a bit?
Basically, it doesn't matter whether we push u to TN and apply dx, or if we pull dx to T*M and apply it to u -- is this just the "d commutes with push-forwards/pullbacks" statement?

I found some info here, it seems to be pretty elementary.

Anyway, thanks again for the help. [/edit]

Last edited:
Every step in my calculation is a simple application of a definition, of either the tensor product or the pullback. If you meant the third equality sign, I'm just using the definition of a pullback there. I suppose I could have dropped some of the parentheses though and wrote the second line as

$$=f^*dx_i(u)f^*dx_j(v)=f^*dx_i\otimes f^*dx_j(u,v)$$

## 1. What is the concept of tensor product commutes with pullback?

The concept of tensor product commutes with pullback is a mathematical property that describes how the tensor product operation interacts with the pullback operation. In simple terms, it means that the order in which these operations are performed does not affect the final result.

## 2. Why is it important to understand tensor product commutes with pullback?

Understanding this property is crucial for many applications in mathematics and physics, particularly in the study of manifolds and vector spaces. It allows for simplification of calculations and provides a deeper understanding of the underlying structures.

## 3. How is tensor product related to pullback?

The tensor product is a mathematical operation that combines two vectors to create a new vector space. The pullback is a transformation that maps a vector from one space to another. The commutativity between these operations means that the pullback can be applied before or after the tensor product without changing the final result.

## 4. What are the conditions for tensor product to commute with pullback?

The tensor product and pullback must be defined on the same vector spaces, and the pullback operation must be linear. Additionally, the tensor product must be associative and compatible with the pullback, meaning that the pullback of a tensor product is equal to the tensor product of the pullbacks.

## 5. Can tensor product commute with pullback in all cases?

No, there are some cases where the tensor product does not commute with the pullback operation. This usually occurs when the vector spaces involved have different dimensions or when the tensor product is not well-defined. It is important to check the conditions for commutativity before assuming it holds true.

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