Bruce Bartlett Baez Munian

Note: Please try to keep the sizes of your pdf scans to a minimum by scanning them in at average resolution. The server barfs at file sizes greater than 100Kb!

Bruce: Congratulations to Peter for going the extra mile in question 9 and showing that $\{\partial_\mu\}$ forms a basis (over $C^\infty (\mathbb{R}^n)$) for the vector fields on $\mathbb{R}^n$.

Bruce: Paul - I’m wondering why you needed to find two disjoint open subsets of $\mathbb{R}^n$ which were both diffeomorphic to $\mathbb{R}^n$ in exercise 6.

Paul: Oh, I think I see what I did wrong: the family of open sets that I specify to cover $M \cup N$ is unnecessarily large because it includes all sets of the form $A \cup B$ instead of just all A’s and all B’s. ($A \subset M$, $B \subset N$.) With such a large family of open sets it could happen that the images of A and B in $\mathbb{R}^n$ overlap and you would need this device to avoid that.

Bruce: I see.

Bruce: By the way, if anyone wants to find out more about the sheaf-theoretic approach to manifolds (‘diffeological spaces’), one place to start is here.

Bruce: Note that there were typos in the questions for 18 and 23 as outlined on the book errata page.

Bruce: Nice Paul!

Bruce: Well done Gerrit, nice answers. Still waiting for numbers 30 and 32, which are the tricky ones!

In exercise 28, the $\lambda$ on the $\partial x_\lambda$ should be a superscript.

In exercise 35, you are a bit naughty to write $dx^\mu$ as being the dual basis to $\partial_\mu$, because in the preceding paragraph it was apriori defined as $dx^\mu := \phi^*(dx^\mu)$ where the $dx^\mu$ on the right is by definition the coordinate 1-forms on \mathbb{R^n} defined on pages 42-43.

In exercise 44, you don’t talk about whether the wedge products are linearly independent or not. (I didn’t prove this either! But I mentioned it. A proof is in the appendix in Looijenga. The idea is to prove things are linearly independent by giving a well-defined functional which separates them.).

Bruce: Paul, Your argument for the first part of exercise 30 is good, more simple than mine. I think there’s a small subtlety in the second part: you have proved that $\mathbb{R^\omega_p$ only depends on $v_p$ for those $v_p$ that are obtained as restrictions of vector fields on $M$, i.e. $v_p = v|_p$; you could mention that this is no real restriction since any tangent vector can be extended to a vector field.

In question 32, you have only really proved uniqueness. The hard part is existence, i.e. you need to explicitly check that $\phi^*(\omega)$ is indeed a smooth 1-form (in your proof, you need to check that indeed $\nu$ ends up in $C^\infty(M)$. The only way I saw of doing this was an explicit local calculation, as in my proof.

Nice solution for 35! Better than mine.

Question 38 was a bit simpler than you made it (see Gerrit’s and my solution!). However, it’s true that the issue of smoothness was swept under the rug (it wasn’t dealt with explictly in the question). Regarding your question on whether $S^\mu_nu (p)$ will be smooth, the answer is: yes, because there is an algebraic smooth formula for the inverse of a matrix (the matrix of cofactors divided by the determinant).

For question 41, well done for comparing to $\vec{u} \cdot (\vec{v} \times {w})$ which Gerrit and I forgot to do.

In exercise 44, it’s good that you worry about whether the wedge products are linearly independent or not. A proof is in the appendix in Looijenga. The idea is to prove things are linearly independent by giving a well-defined functional which separates them.

Paul: Thank you for your comments Bruce.

In question 30: You’re right, that should be mentioned. And now that you mention it, I think they assume this in the definition of the cotangent vector on the previous page, since otherwise it is not defined on the entire tangent space.

In the same question, I just found two typos in my answer: $\omega_p(v)$ should have been $\omega_p(v_p)$, and similarly for the next line.

In question 38: Indeed, your solution is much simpler. But I think it shows that $T_{\mu}^{\nu}(p)$ is invertible if and only if the tangent vectors $e_{\mu,p}$ form a basis for $T_p$. So then you would still need to show that the vector fields $e_{\mu}$ are a basis of vector fields on $U$ if and only if the induced tangent vectors are a basis for each tangent space. The forward direction follows because any tangent vector is the restriction of some vector field, but the reverse direction seems more complicated, and needs to use the fact that the functions $T_{\mu}^{\nu}$ are smooth on $U$. (I think their smoothness follows immediately from the statement that the $e_{\mu}$ are vector fields.)

Bruce: Regarding question 38: I agree with you. Let’s phrase the question more precisely, in the way I’m sure Baez intended: “Suppose $T^\nu_\mu$ are smooth functions on $U$, and set $e_\mu = T^\nu_\mu \partial_\nu$. Show that…”. On the other hand, it’s also true, as you say, that if had been told explicitly that the $e_\mu$ are vector fields, then it necessarily follows that the functions $T^\nu_\mu$ must be smooth. That’s because the functions $T^\nu_\mu$ are, as a matrix, simply the columns of $e_\mu$ expressed in terms of $\partial_\mu$, which we are told are smooth functions.

Bruce: Solomon, in exercise 50, there is a subtlety about global vs local which you should address explicitly, as well as uniqueness.

In exercise 52, you should state that this is only true in finite dimensions (this is omitted from the question). Happy with exercise 53, although you only need to check it on basis vector fields $e_\beta$ (i.e. you don’t need to check it on $u^\beta e_\beta$). Also for 54.

In exercise 69, the gist of your argument is correct, but you should be a bit clearer on some points. In the second line of your calculation, you have $e^{j_1} \wedge \cdots \wedge e^{j_{n-p}}$ which is a bit confusing, since there are no $j$‘s appearing in the rest of the expression. It is the complement, as you say, but the ordering is a bit confusing, because in the paragraph you say $i_1 < i_2 < \cdots < i_p$, which isn’t true - there is no ordering on the sum $\omega = \frac{1}{p!} \omega_{i_1 \cdots i_p} e^{i_1} \wedge \cdots \wedge e^{i_p}$.

Bruce: Gerrit, I’ll go through your answers one by one. Number 46, good.

Number 47, be careful here, there is a subtle distinction to be made. Baez has defined a differential $p$-form as an abstract linear combination of wedge products like $\omega_1 \wedge \cdots \wedge \omega_p$. Although each $\omega_i$ is concretely a map from $Vect(M)$ to $C^\infty(M)$, he has not given a concrete interpretation to wedge products like $\omega_1 \wedge \cdots \wedge \omega_p$. In your answer, you have thought of p-forms as being multilinear alternating functionals in their own right, i.e. you are using the identification $\wedge^p V^* \cong (\wedge^p V)^*$. You need to explicitly write down the precise identification you are using here, since there are two conventions (one can decide to divide by $p!$ or not).

Numbers 48 and 49, good.

In number 50, you need to add one line at the end. Namely, you have proved that locally one can write $F = B + E \wedge dt$. You need to argue that, therefore, one can write it globally in this way.

Number 51, good. Number 52, good, but you need to remark that this argument only works in finite dimensions (this subtlety was left out of the question. See my comment to Solomon’s answer above). Numbers 53-60, good.

Number 61, nice answer. Number 62, good answer (looks like taken from Bott and Tu?).

Number 63, good, but you need to make clear that this cosntruction will only construct a coordinate system $y_i$ such that $dy_i = e_i$ at $p$. In general, we can’t ensure that $dy_i = e_i$ everywhere on the open set $U$: that amounts to solving a certain partial differential equation, and there are many examples where it can’t be done.

Numbers 64-69, good. Numbers 70-71, good, well written out.

Gerrit:

No. 47, thanks for the comments. I see what you mean. A silly oversight on my part.

No. 50, I should have added that defining a form on all the charts is equivalent to defining it on the whole manifold.

No. 62, Indeed I did obtain the solution of 62 from the book. You caught me! I did go through it carefully though and I understand how it works.

No. 63, thanks understood.

Bruce: Gerrit, you have made a double sign error in 75. When you take $d_S \vec{E} e^{ik_\mu x^\mu}$, you must realize that $\vec{E}$ is a 1-form, so when you use the Leibniz law to “jump” the $d$ over the $\vec{E}$, you pick up a minus sign. Also, ${}^3 k \wedge \vec{E}$ = -$\vec{E} \wedge {}^3 k$ since they are one-forms.

Number 76 is a bit messy but not so bad, that’s the right idea, but you could do it more elegantly by writing out the vectors in terms of their magnitudes and unit vectors, i.e. $\vec{E} = |\vec{E}| \hat{E}$ and so on. It amounts to the same thing. Otherwise good.

Bruce: I didn’t answer 89 in the way the book intended. The correct answer, as in Peter, Paul and Gerrit, is that one ‘should be taking $\infty$ into account’ as $\int_0^\infty f'(x) dx$ could diverge while $-f(0)$ will always be a finite number.

Bruce: Folks, observe how Gerrit did 91. To show that $S^{n-1} \subset \mathhbb{R}^n$ is a submanifold, one can use the nifty implicit function theorem which tells us that if your aspirant submanifold $S$ is the zero set of a smooth function $f : \mathbbb{R^n} \rightarrow \mathbb{R}$ whose derivative $Df$ is surjective everywhere on $S$, then $S$ is indeed a smooth submanifold of \mathbb{R^n}. The idea is simple: if the derivative is surjective, there must be some directions in which the function $\mathbb{R^f$ changes. So by linear algebra, there will be a complimentary subspace of directions in which the function $f$ doesn’t change; this will be the tangent space of our submanifold. See example 2.6 on page 12 of Looijenga’s notes.

Bruce: Paul, in 86, the orientation comes in not so much at the global level but rather at the level of each chart $(U, \phi)$. We need to be able to convert the ‘abstract differential form’ integral $\int_\phi(U) f(x) dx^1 \wedge \cdots \wedge dx^n$ into a concrete second-year calculus integral $\int_{\phi(U)} f(x) dx_1 \cdots dx_n$. Notice that in our second year calculus integrals we always counted the parameters from smallest to biggest. This had the effect that our change of variables formula in ordinary calculus features the absolute value of the Jacobian and not the Jacobian itself (see eg. the following 2-variable formula in Stewart pg 1052):

The trouble is that differential forms don’t transform in quite the right way: if you change coordinates, they pick up the Jacobian factor, without the absolute value!

So in order for us to translate our abstract differential form integrals into concrete second-year calculus integrals in $\mathbb{R}^n$, we need an orientation. That way, the determinant above is always positive. In 88, the different signs follow from ‘The Convention on the Orientation of the Boundary’ as follows. The convention is that an ordered basis $B$ is an oriented basis for $T_p \partial M$ iff $v \union B$ is an oriented basis for $T_p M$, where $v$ is an outward pointing vector. When $M$ is one-dimensional, like an interval, this ultimately means that the point $b \in \partial [a,b]$ will have $+$ orientation and the point $a$ will have $-$ orientation.

Paul: Thanks for the clarifications.

Bruce: Folks, Paul did exercise 96 in a nice way. He realized that we might as well assume that $S$ lies in the $xy$ plane. Then the calculations simplify a bit, and it’s easier to see that you recover the ordinary Stokes’ theorem in $\mathbb{R}^3$.

Bruce: Paul, indeed 103 isn’t so relevant. In 105, why did you break it up into two integrals…? In 106, your calculation in the large $r$ regime is fine, it wasn’t quite clear what they wanted, but it also works exactly for all $r$, since $*E = \frac{q}{4 \pi} \sin \phi d \phi \wedge d \theta$ is true for all $r$. (I agree with your remark about what $\phi$ and $\theta$ mean). I agree with your argument in 107 (same as mine). In 110, you are right; also to show it is exact we simply need to calculate some or other integral and show it is nonzero. Indeed, any integral of a hypersurface wrapping around the origin will give the same result, namely the surface area of a hypersphere… not so easy to calculate exactly (see wikipedia), but at least we can easily see it’s nonzero.

Paul: In 104, it’s interesting that you can derive the function $e(r)$ up to a constant. Point taken in 105. I think I forgot a basic rule of integration (that the integral from $a$ to $b$ is defined even if $a$ is greater than $b$). And I was not applying the fundamental theorem of calculus correctly… I assumed you needed an extra constant term. Also, your way of showing that it’s not exact in 110 is simpler than I’d have thought.

Bruce: Paul, in question 3, in your last displayed equation, I think you mean $\langle \nu, \omega \rangle = \langle A^{-1} \nu, A^{-1} \omega\rangle$. In exercise 4, you have a nice quick-and-dirty proof for why the special linear group is indeed a manifold (using the $1 + \epsilon$ in the upper left hand corner trick.) You are on the right track for the orthogonal group, just realize that the map $A \mapsto A^T A$ really takes values not in all matrices, but rather in symmetric matrices. Then the dimensions work out.

Bruce: I thought it would be good to compare the three solutions submitted so far (Huygens’, Paul’s and mine) to question 6 (“Prove that an orthogonal three-dimensional matrix is either a rotation or a rotation followed by a reflection”). Huygens used eigenvalues and Jordan normal form. Paul used a more geometric argument and didn’t used the concept of eigenvector, although it was implicit (his vector “$y$”, the rotation axis, was the eigenvector). I used eigenvectors and orthogonal decomposition to reduce it to the 2x2 case, similar to Huygens.

Bruce: It is good to contrast Paul’s solution of number 15 with my own. Paul worked locally in the sense that he realized that the group multiplication means that everything is determined by what happens in a neighbourhood of 1. I worked infinitesimally in the sense that in fact, everything is determined by what happens at the level of the tangent space at 1, in other words at the level of the Lie algebra.

Bruce: On page 189 we find the following statement:

“Using some topology, it follows that any element of the identity component of $G$ is the product of elements of the form exp($x$).”

This is a great exercise! Try to do it. I included the solution in the first part of my exercises. Here’s a concrete puzzle: show that the matrix

cannot be written as exp($X$) for $X$ in the Lie algebra of $SL(2, \mathbb{C})$!

Bruce: The most elegant approach to prove that, if $E$ is a vector bundle, then $E^*$, $\Lambda^k E$ and so on are also vector bundles in a natural way, is the approach of Atyah in the opening pages of his K-theory book. Namely, one just needs to check that those constructions are functorial and smooth. For instance, the process of taking the $k$th exterior power of a vector space $V$ can be interpreted as a functor from the category of finite-dimensional vector spaces to itself. Moreover, at the level of hom-sets, the the functor is a smooth map $Hom(V, W) \rightarrow Hom (\Lambda^k V, \Lambda^k W)$ since it is just a polynomial in the matrix entries. Then by the arguments in Atiyah (and outlined in my exercises) it follows there will be a natural smooth structure on $\Lambda^k E$ too.

Huygens (5 November 2010): My question may be off topic but it may also be related to the first beginning. How do we get a topological condition for the existence of a Lorentzian metric on a smooth manifold ? For compact one we just need the Euler characteristic to be zero but how about the non-compact case ?

Bruce (8 November 2010): Huygens, I wasn’t aware of there being obstructions to having a Lorentzian metric. I had assumed the partition of unity would do the trick. I will think about that! In the meantime, on wikipedia it appears to say that there is no obstruction to admitting a Lorentzian metric if the manifold is non-compact.