# Having Trouble with Analysis

1. Mar 20, 2005

### 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.

Last edited: Mar 20, 2005
2. Mar 20, 2005

### 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.

$$= \int _{C_0} d\omega[/itex] By Stokes' Theorem [tex] = \int _{C_0} 0$$

Since $\omega$ is closed

$$= 0$$

Therefore:

$$\int _{f - g} \omega = 0$$

$$\int _f \omega - \int _g \omega = 0$$

$$\int _f \omega = \int _g \omega$$

as required.

Last edited: Mar 20, 2005
3. Mar 20, 2005

### 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.

4. Mar 20, 2005

### 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)...

5. Mar 20, 2005

### 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 $J = f'(p)$, 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?

Last edited: Mar 20, 2005