# Proving Pushforward Product Isomorphism: M1 x ... x Mk to M1 + ... + Mk

• robforsub
In summary, the conversation discusses the map a which is defined as a linear isomorphism between the tangent space at the product manifold M1XM2X...XMk and the direct sum of tangent spaces at each individual manifold. The participants are trying to prove that a is a bijection by taking a natural basis in the image and finding its pre-image, and also considering the manifold structure on M x N. They suggest experimenting with basis associated with different charts to prove the statement.
robforsub
Let M1,..,Mk be smooth manifolds, and let Pi_j be the projection from M1XM2X...XMk->Mj. Show that the map a:T_(p1,...,pk)(M1XM2X...Mk)->T_p1(M1)$$\oplus$$...$$\oplus$$T_pk(Mk)

a(X)=(Pi_1*X,Pi_2*X,...,Pi_k*X) is an isomorphism.

The way I am thinking to prove the statement is to show that a is a bijection, since a is already a linear map. And I have no clue how to show it, help needed!

I would take a natural basis in the image and "guess" what would be its pre-image - then check.

But, don't we need to take component functions of arbitrary X into account?

Recall the manifold structure on M x N. If (U,f) is a chart of M^m around p and (V,g) is a chart of N^n around q, then (U x V, f x g) is a chart of M x N around (p,q). If ∂/∂x^i is the basis of T_pM associated with the chart (U,f) and ∂/∂y^i is the basis of T_qN associated with the chart (V,g), call (∂/∂x'^1,..., ∂/∂x'^m, ∂/∂y'^1,..., ∂/∂y'^n) the basis associated with the chart (U x V, f x g).

Just doing this should give you some room for your experiments.

I can offer some guidance on how to approach this problem. First, it is important to understand the definitions involved. The pushforward map, denoted by a, is a linear map that takes tangent vectors from the product manifold M1XM2X...XMk and maps them to tangent vectors on the individual manifolds M1, M2, ..., Mk. The notation T_(p1,...,pk)(M1XM2X...Mk) represents the tangent space at a point (p1,...,pk) on the product manifold, while T_p1(M1)\oplus...\oplusT_pk(Mk) represents the direct sum of the tangent spaces at points p1, ..., pk on the individual manifolds.

To show that a is an isomorphism, we need to prove that it is both injective and surjective. In other words, we need to show that a is a one-to-one mapping and that it covers the entire target space.

To prove injectivity, we can assume that a(X)=0, where X is a tangent vector on the product manifold. This means that all the components of a(X) must be zero, i.e. Pi_j*X=0 for all j=1,...,k. Since the projections Pi_j are onto, this implies that X=0, proving that a is injective.

To prove surjectivity, we need to show that for any tangent vector (v1,...,vk) in T_p1(M1)\oplus...\oplusT_pk(Mk), there exists a tangent vector X in T_(p1,...,pk)(M1XM2X...Mk) such that a(X)=(v1,...,vk). This can be achieved by choosing X=(v1,v2,...,vk), where vj is the projection of (v1,...,vk) onto T_pj(Mj).

Thus, we have shown that a is both injective and surjective, and therefore it is an isomorphism. This proves the statement that the pushforward map a is an isomorphism from the product manifold M1XM2X...XMk to the direct sum of tangent spaces T_p1(M1)\oplus...\oplusT_pk(Mk).

## 1. What is the pushforward product isomorphism?

The pushforward product isomorphism is a mathematical concept in differential geometry that describes the relationship between different manifolds. It is a type of linear transformation that maps the tangent spaces of a product manifold to the tangent space of the individual manifolds.

## 2. How is the pushforward product isomorphism used in proving M1 x ... x Mk to M1 + ... + Mk?

To prove that the pushforward product isomorphism holds for M1 x ... x Mk to M1 + ... + Mk, we use the fact that the pushforward transformation preserves the algebraic structure of the tangent spaces. This allows us to show that the pushforward map is an isomorphism between the two manifolds.

## 3. Are there any limitations to the pushforward product isomorphism?

Like any mathematical concept, the pushforward product isomorphism has its limitations. It only applies to smooth manifolds and cannot be extended to non-smooth manifolds. Additionally, the pushforward map must be bijective in order for the isomorphism to hold.

## 4. Can the pushforward product isomorphism be extended to higher dimensions?

Yes, the pushforward product isomorphism can be extended to higher dimensions. In fact, it is a fundamental concept in differential geometry and is used to prove many important theorems and results.

## 5. How is the pushforward product isomorphism related to other mathematical concepts?

The pushforward product isomorphism has connections to other mathematical concepts such as the inverse function theorem and the chain rule. It is also related to the concept of a tangent bundle, which is used to study the behavior of a manifold near a point.

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