Homepage Solution manuals John Lee Introduction to Smooth Manifolds Problem 3-2 (The tangent space to a product manifold)

Problem 3-2 (The tangent space to a product manifold)

Proposition 3.14 (The Tangent Space to a Product Manifold). Let M 1 , , M k be smooth manifolds, and for each j , let π j : M 1 × × M k M j be the projection onto the M j factor. For any point p = ( p 1 , , p k ) M 1 × × M k , the map

α : T p ( M 1 × × M k ) T p 1 M 1 T p k M k

defined by

α ( v ) = ( d ( π 1 ) p ( v ) , , d ( π k ) p ( v ) )

is an isomorphism. The same is true if one of the spaces M i is a smooth manifold with boundary.

Answers

Linearity. First, we demonstrate that α is a linear map. Obviously, we use the fact that each component d ( π j ) p in itself is linear (cf. Proposition 3.6(a)). Let v , w T p ( M 1 × × M k ) and c be arbitrary. We then have

α ( cv + w ) = ( d ( π 1 ) p ( cv + w ) , d ( π k ) p ( cv + w ) ) = ( cd ( π 1 ) p ( v ) + d ( π 1 ) p ( w ) , cd ( π k ) p ( v ) + d ( π k ) p ( w ) ) = c ( d ( π 1 ) p ( v ) , , d ( π k ) p ( v ) ) + ( d ( π 1 ) p ( w ) , , d ( π k ) p ( w ) ) = c α ( v ) + α ( w )

Bijectivity. Now we demonstrate that α is bijective. To do so, we explicitly provide an inverse map by reverse engineering α . For each j , let ι j : M j M 1 × × M k be the embedding onto the M j factor of p , i.e., ι j ( x ) = ( p 1 , , p j 1 , x , p j + 1 , , p k ) . For any point ( p 1 , , p k ) M 1 × × M k , the map

β : T p 1 M 1 T p k M k T p ( M 1 × × M k )

defined by

β ( w 1 w k ) = d ( ι 1 ) p ( w 1 ) + + d ( ι k ) p ( w k )
(1)

is the inverse for α . To see why, consider some v 1 v k T p 1 M 1 T p k M k . Unwrapping the definitions and using various properties from Proposition 3.6, we obtain

( α β ) ( v 1 v k ) = α ( j = 1 k d ( ι j ) p ( v j ) ) = j = 1 k α ( d ( ι j ) p ( v j ) ) = j = 1 k ( d ( π 1 ) p ( d ( ι j ) p ( v j ) ) , , d ( π k ) p ( d ( ι j ) p ( v j ) ) ) = j = 1 k ( d ( π 1 ι j ) p ( v j ) , , d ( π k ι j ) p ( v j ) )

Notice that the functions themselves satisfy ( π i ι j ) ( p j ) = π i ( ι j ( p j ) ) = π i ( p ) = p i . In other words, π i ι j is the identity function for i = j and the constant function for i j . By Proposition 3.6, d ( π j ι j ) = d ( Id T p j M j ) and by Problem 3.1, d ( π i ι j ) = 0 for i j . We get

( α β ) ( v 1 v k ) = j = 1 k ( 0 , , 0 , Id T p j M j ( v j ) , 0 , , 0 ) = j = 1 k ( 0 , , 0 , v j , 0 , , 0 ) = ( v 1 , , v k )

In other words, α β is the identity function on T p 1 M 1 T p k M k . Thus, α is surjective. Since α is a linear transformation and since T p ( M 1 × × M k ) and T p 1 M 1 T p k M k have the same dimension1, α must be a bijection.

Manifolds with boundary. Suppose that M j is a smooth manifold with boundary for some j . Then M 1 × × M k is a smooth manifold with boundary as well. Therefore, it has a tangent space at each point with well-defined differentials. We can therefore apply the same proof to the case of a manifold with boundary.

1cf. Example 1.34 and Proposition 3.10/3.12 and the fact that a tangent space is a manifold. Both spaces are thus of dimension i = 1 k dim ( M i ) .

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2023-08-30 15:46
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