What is the proof for the existence of a diffeomorphism between A and B?

[AKG]

Here's a bunch of problems, some of which I think I've done right, others I've attempted, others I have no clue. Any help would be appreciated, and please let me know if the one's I've done are right:

1. Let M_1 \subset \mathbb{R}^{n_1},\ M_2 \subset \mathbb{R}^{n_2 be oriented manifolds without boundary, of dimensions k_1 and k_2 respectively.

Show M_1 \times M_2 is a manifold without boundary in \mathbb{R}^{n_1 + n_2} with a natural orientation induced by M_1 and M_2

Well, \forall \, x_1 \in M_1, there is an open set U_{x_1} containg x_1 and an open set V_{x_1} \subset \mathbb{R}^{n_1}, and a diffeomorphism h_{x_1} : U_{x_1} \to V_{x_1} such that

h_{x_1}(U_{x_1} \cap M_1) = V_{x_1} \cap (\mathbb{R}^{k_1} \times \{ 0 \}), and something similar for points in M_2. Let x = (x_1,\, x_2) be any point in M_1 \times M_2. Define a function h_x : U \to V where U = U_{x_1} \times U_{x_2} by:

h_x(u_1, u_2) = (h_{x_1}(u_1), h_{x_2}(u_2))

Then define the permutation p by:

p(y_1, \dots , y_{n_1 + n_2}) = (y_1,\dots ,y_{k_1}, y_{n_1 + 1} ,\dots, y_{n_1 + k_2}, y_{k_1 + 1}, \dots , y_{n_1}, y_{n_1 + k_2 + 1}, \dots , y_{n_1 + n_2})

Then define H_x = p \circ h_x for each x. This function satisifies the conditions required to make M_1 \times M_2 a manifold. (Is it right?)

2. Let S be the set defined by the equations:

x^2 + y^2 + z^4 = 3,\ x^3 - y^3 + z(1 + xy) = 2

Let f(x, y, z) = e^{x + yz} + x^3y

Show that, for P = (1, 1, 1) and some \epsilon > 0, M = S \cap \mathcal{B}_P (\epsilon ) is a manifold, where \mathcal{B}_P (\epsilon ) is the open ball of radius \epsilon centered at point P.

Don't really know how to do this one.

 

[AKG]

3. Let M = {(x, y, z, w) | x² + y² = z² + w² = 1/2}. Show that M is a manifold

Let W \subset \mathbb{R}^2 be the set (0,\, 2\pi ) \times (0,\, 2\pi ) and define a function f : U --> M by:

f(\theta ,\phi ) = \frac{1}{\sqrt{2}}(\cos \theta , \sin \theta , \cos \phi , \sin \phi )

I believe, then, that f is 1-1, differentiable, with f(W) = M, f'(y) has rank 2 for each y in W, and the inverse of f is continuous. Therefore, M is a manifold.

4. Let U \subset \mathbb{R}^n be any open set. Let:

f : [0, 1] \to U,\ \ g : [0, 1] \to U

be to C^{\infty} curves. If there is a C^{\infty} function

F:[0,1] \times [0,1] \to U

such that

F(0, t) = f(t),\ F(1, t) = g(t),\ f(0) = g(0) = F(s, 0),\ f(1) = g(1) = F(s, 1)

for all s, t in [0, 1], we say that f and g are homotopic (and F is a homotopy between them). Show that if f and g are homotopic, then

\int _f \omega = \int _g \omega

for any 1-form on U using the following steps

(a)Show f - g = \partial C_0 + C_1 + C_2 where C_1,\ C_2 are two degenerate 1-chains and C_0 is a 2-cube.

Okay, we have:

\partial (-F) (t) = \sum _{i = 1} ^2 \sum _{\alpha = 0} ^1 (-1)^{i + \alpha}(-F)_{(i, \alpha)}(t)

= -(-F)(0, t) + (-F)(1, t) + (-F)(t, 0) - (-F)(t, 1)

= F(0, t) - F(1, t) - F(t, 0) + F(t, 1)

= f(t) - g(t) - f(0) + f(1)

-F is a 2-cube since it's a C^{\infty} function whose domain is [0,1]^2. Let C_0 = -F, so we have:

\partial C_0 = f - g - f_0 + f_1

where f_{\alpha} is the 1-chain defined by f_{\alpha}(t) = f(\alpha ). We then get:

f - g = \partial C_0 + f_0 - f_1 = \partial C_0 + f_0 + (-f_1)

If we let C_1 = f_0 and C_2 = -f_1, we're done.

(b) Justify the following inequalities, if \omega is closed:

\int _{f - g}\omega = \int _{\partial C_0} \omega = \int _{C_0} d\omega = 0

Simple enough, I think:

\int _{f - g} \omega = \int _{\partial C_0 + C_1 + C_2} \omega

= \int _{\partial C_0} \omega + \int _{C_1} \omega + \int _{C_2} \omega

= \int _{\partial C_0} \omega

Since C_1,\ C_2 are constant functions.

= \int _{C_0} d\omega[/itex]<br /> <br /> By Stokes&#039; Theorem<br /> <br /> = \int _{C_0} 0<br /> <br /> Since \omega is closed<br /> <br /> = 0<br /> <br /> Therefore:<br /> <br /> \int _{f - g} \omega = 0<br /> <br /> \int _f \omega - \int _g \omega = 0<br /> <br /> \int _f \omega = \int _g \omega<br /> <br /> as required.

 

[AKG]

5. We define an open set A \subset \mathbb{R}^n to be cohomologically trivial if any closed form on A is exact. Show that if A and B are diffeomorphic (i.e. there exists a diffemorphism f : A --> B), then A is cohomologically trivial if and only if B is.

I have little idea how to do this, my only guess is that it has something to do with Poincaré's Lemma.

6. Let V be an n-dimensional vector space over the reals. Let \mathcal{T}^k(V) be the space of k-tensors on V, and \wedge ^k(V) the space of alternating k-tensors, where k = 0, ..., n.

(a) Let f : V \to V be a linear transformation. Show that f* : \mathcal{T}^k(V) \to \mathcal{T}^k (V) and f* : \wedge ^k(V) \to \wedge ^k (V) are isomorphisms for all k = 0, ..., n if and only if f is an isomorphism.

Suppose S and T are k-tensors with f*(S) = f*(T). Then:

f*(S)(v_1, \dots, v_k) = f*(T)(v_1, \dots, v_k)

S(f(v_1), \dots, f(v_k)) = T(f(v_1), \dots, f(v_k))

(S - T)(f(v_1), \dots, f(v_k)) = 0

Since this is true for any choice of the v_i, and since f is an isomorphism, we have that (S - T)((f(V))^k) = (S - T)(V^k)\{ 0 \}, therefore, (S - T) must be the zero tensor, therefore S = T. This implies that f* is 1 to 1, and since it is linear, this means that it is an isomorphism. Now, if f were not an isomorphism, say its range were W, some subspace of V. Then if f*S = f*T, all we would have is that:

(S - T)((f(V))^k) = (S - T)(W^k) = \{ 0 \}

But if S and T could be any tensors that "behaved" that way on W_k, then S need not equal T, i.e. there could be distinct tensors S and T that did the same things to elements of W^k but differed in their "behaviour" on other vectors, and hence f* would not be 1-1, and hence not an isomorphism.

(b) Let T be a symmetric, positive definite 2-form on V. Let v_1, \dots , v_k \in V be vectors and let \phi _1 , \dots , \phi _k \in V* be the elements defined by:

\phi _i (w) = T(v_i, w)

Show that \phi _1 \wedge \dots \wedge \phi _k \neq 0 if and only if the v_i are linearly independent.

Now, if they are linearly dependant, then we can write:

v_1 = \sum _{i = 2} ^k a_iv_i

So:

\phi 1 (w) = T(\sum _{i = 2} ^k a_iv_i, w) = \sum _{i = 2} ^k a_i\phi _i(w)

In this case, we get:

\phi _1 \wedge \dots \wedge \phi _k = \left ( \sum _{i = 2} ^k a_i\phi _i \right ) \wedge \phi _2 \wedge \dots \wedge \phi _k

= \sum _{i = 2} ^k a_i\phi _i \wedge \phi _2 \wedge \dots \wedge \phi _i \wedge \dots \wedge \phi _k

= -\sum _{i = 2} ^k a_i\phi _i \wedge \phi _2 \wedge \dots \wedge \phi _i \wedge \dots \wedge \phi _k

... by switching the \phi _i with the \phi _i. Therefore, since the product is equal to its own negative, it is zero, which is what we wanted to show. Now, I'm not entirely sure how to prove the other way. So far, if we define

\psi _i (x) = T(w_i, x)

Then I can show that:

\phi _1 \wedge \dots \wedge \phi _k (w_1, \dots, w_k) = \psi _1 \wedge \dots \wedge \psi _k (v_1, \dots, v_k)

Now, if the wedge product of the \phi _i is zero, then when evaluated at any (w_1, \dots , w_k), it will give zero, which means that \psi _1 \wedge \dots \wedge \psi _k (v_1, \dots, v_k) = 0 for any choice of \psi _i, in other words, if \omega \in \wedge ^k (V), then \omega (v_1, \dots , v_k) = 0. But if the v are independent, let their span be the k-dimensional vector space W. So if we take the appropriate restriction of \omega, we'd essentially have that every \omega \in \wedge ^k (W) is the zero tensor, which is impossible, hence the v would have to be dependent.

 

[AKG]

Last one

7. Let T be defined by

T = \{ (x, y, z, w) \in \mathbb{R}^4 : x^2 + y^2 = 1,\ z^2 + w^2 = 1\}

(a)Show that

f : (0, 2\pi ) \times (0, 2\pi ) \to T

given by

f(\theta , \phi ) = (\cos \theta , \sin \theta, \cos \phi , \sin \phi )

is a co-ordinate system.

Well, f is obviously differentiable, and its domain (let's call it W) is obviously open. To show that it is 1-1, it suffices to show that g(x) = (cos(x), sin(x)) is 1-1. Suppose g(x) = g(y), then:

cos(x) = cos(y), sin(x) = sin(y)

Therefore, since cos(x) = cos(y), either x = y or x = 2п - y. If x = y, we're done, otherwise:

sin(x) = sin(y)
sin(2п - y) = sin(y)
-sin(y) = sin(y)
y = п, therefore x = 2п - п = п = y. Therefore, g is 1-1, and hence so is f.

I can also show that f(W) = T - {(1, 0, 1, 0)}, which can be done by showing that g((0, 2п)) = {(x, y) : x² + y² = 1} - {(1, 0)}. Any point (x, y) can be written as (r*cos(t), r*sin(t)), but we know r = 1, so every point (x, y) in this set can be written as (cos(t), sin(t)) for some t in (0, 2п)... actually, this seems to give me problems with the point (1, 0, 1, 0). Anyways, its easy to show that f'(y) has rank 2 for all y in W, since f'(y) takes the form:

(a 0)
(b 0)
(0 c)
(0 d)

And at least one of c and d must be non-zero, and at least one of a and b must be non-zero, so we have at least the rows (a 0) and (0 c), without loss of generality, which are clearly linearly independent. The inverse of f needs to be continuous, but I believe inverse function theorem guarantees this. However, what do I need to show to claim that the inverse function theorem applies?

(b) Let T have the orientation \mu such that f is positively oriented with respec to \mu. Show that for any 2-form \omega defiend on a neighborhood of T,

\int _T \omega = \int _{(0, 2\pi)\times (0, 2\pi )}f*\omega

I'm not sure how to do this, but I also have the feeling that I don't need to know it (for the test I'm studying for), but I've put it in just in case, and some part of it is necessary for the next question.

(c) Integrate the 2-form \omega = yw\, dx \wedge dz over T given the orientation \mu as in part (b).

\int _T \omega = \int _{(0, 2\pi)\times (0, 2\pi )}f*\omega

= \int _0 ^{2\pi} \int _0 ^{2\pi} f*(yw\, dx \wedge dz)

= \int _0 ^{2\pi} \int _0 ^{2\pi} (yw \circ f)f*(dx \wedge dz)

= \int _0 ^{2\pi} \int _0 ^{2\pi} (y \circ f \cdot w \circ f)f*(dx) \wedge f*(dz)

= \int _0 ^{2\pi} \int _0 ^{2\pi} (\sin \theta \sin \phi)\left (\frac{\partial \cos \theta}{\partial \theta} d\theta + \frac{\partial \cos \theta}{\partial \phi} d\phi \right ) \wedge \left (\frac{\partial \cos \phi}{\partial \theta} d\theta + \frac{\partial \cos \phi}{\partial \phi} d\phi \right )

= \int _0 ^{2\pi} \int _0 ^{2\pi} (\sin \theta \sin \phi)\left (\frac{\partial \cos \theta}{\partial \theta} d\theta \right ) \wedge \left (\frac{\partial \cos \phi}{\partial \phi} d\phi \right )

= \int _0 ^{2\pi} \int _0 ^{2\pi} (\sin \theta \sin \phi)(- \sin \theta d\theta ) \wedge (- \sin \phi d\phi )

= \int _0 ^{2\pi} \int _0 ^{2\pi} (\sin ^2 \theta \sin ^2 \phi)d\theta \wedge d\phi

= \int _0 ^{2\pi} \int _0 ^{2\pi} \sin ^2 \theta \sin ^2 \phi d\theta d\phi

= \left ( \int _0 ^{2\pi} \sin ^2 \theta d\theta \right )^2 = \pi ^2

Now, we actually haven't done integration on manifolds, so this might not be right. For one, I made no use of the orientation. However it looks like regular integration (or at least integration on chains)...

 

[AKG]

I notice that I did #3 wrong, since f(W) is not M, there are "problems" at the point 1/√(2)(1,0,1,0). However, I was trying to find an f that would work for all x in M, but I only need an f to work in an open set for each x in M. The given f will work for most x except that problematic one. For that point, I can easily choose a different open set: instead of (0, 2π) x (0, 2π), I can take (-π/2, π/2)².

Related to this is one of the problems I had with question 7, which I underlined. For f to be a co-ordinate system, then for all points x in M, there must be an open set containing x such that f((0, 2π)²) = M ∩U. But in fact there is no point y in (0, 2π)² such that f(y) = (1, 0, 1, 0), which is a point in M. So it seems f is not a co-ordinate system.

So, I feel okay about most of the problems so far. If anyone could tell me if I've done 1, 3, 4, 6, and 7 right, that would be very much appreciated. However, I am still pretty lost on questions 2 and 5. Any ideas?

For question 2, I'd like to be able to at least picture this manifold. I can't even tell if its supposed to be 1-dimensional or 2. For question 5, it seems to me that I only need to prove one way, and the other way will follow since if there is a diffeomorphism from A to B, then there is a diffeomorphism, f^{-1}, from B to A.

Well, a closed k-form on A maps points in A to the space of alternating k-tensors on the tangent space of A at the point. Now, the tangent space of A at the point p is isomorphic to the tangent space of B at the point f(p), since they are both isomorphic to \mathbb{R}^n. Let \omega be a k-form on A, let p be a point in A, then:

\omega (p) \in \wedge ^k (A_p)

where A_p is the tangent space to A at p. If \eta is a k-form on B, then:

\eta (f(p)) \in \wedge ^k (B_{f(p)})

Let v_1, \dots , v_k be vectors in \mathbb{R}^n, and let (v_i)_x be a vector in the tangent space to the point x. Then:

\omega (p)((v_1)_p, \dots , (v_k)_p) = \omega (f^{-1}(f(p)))((v_1)_{f^{-1}(f(p))}, \dots , (v_k)_{f^{-1}(f(p))})

Let J = f&#039;(p), then the above is:

\omega (f^{-1}(f(p)))((J^{-1}Jv_1)_{f^{-1}(f(p))}, \dots , (J^{-1}Jv_k)_{f^{-1}(f(p))})

= (f^{-1})^*\omega (f(p))((Jv_1)_{f(p)}, \dots , (Jv_k)_{f(p)})

Let's call (f^{-1})^*\omega = \eta _1, and f(p) = q, then the above is:

\eta _1 (q) ((Jv_1)_q, \dots , (Jv_k)_q)

Clearly, q is in B, and \eta _1 is a k-form on B. J, since it is the inverse of a matrix, is invertible, thus has rank n, and is thus an isomorphism. If \omega is closed, then for all p in A, and for all v_1, \dots , v_k, d\omega is zero. But if this is true for all p in A, then the above expression with \eta _1 is zero for all q in B, since f(A) = B (f is a diffeomorphism, hence invertible, hence onto), and for every Jv_1, \dots , Jv_k, which is really just every v_1, \dots , \v_k since J is an isomorphism. Therefore, \eta _1 is closed.

(... I'm really not sure where I'm going with all of this ...)

So it seems every closed k-form on A can be expressed in terms of a closed k-form on B. If every k-form on B is exact, then every form like \eta _1 can be expressed as dH _1 for some k-1 form H _1. But then:

d((f^{-1})^*H _1) = (f^{-1})^*(dH _1)

by a theorem in my book, then giving:

= ((f^{-1})^*)\eta _1 = \omega

So \omega is exact, as required. I have a feeling this isn't entirely right... Any ideas?