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

[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!

 

[arkajad]

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

 

[huyichen]

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

 

[quasar987]

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.

 

[blue_raver22]

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).