@Danu: I have a definition that a smooth vector field is called a Killing field if its local flow is an isometry onto its image. Is that what you mean?
Then the "proof" Carroll gives is: 1) Go to locally inertial coordinates. 2) Observe that the Riemann tensor should be invariant under Lorentz transformations because of maximal symmetry 3) Recall that only the Kronecker delta, the metric and the Levi-Civita tensor have this property 4) Try to replicate the symmetries of the Riemann tensor 5) Find the above form
@Huy Sounds about right. The way physicists think about it is: The Lie derivative of the metric w.r.t. a Killing vector vanishes
@Huy It's the one physicists use. I just completed a (real mathematics) course in Riemannian geometry, but we spent all semester doing Chern-Weil theory instead of actual Riemannian geometry so we didn't get to Killing fields :(
I'm doing a mathematics course in Riemannian geometry too and unfortunately the prof hardly does a single proof and leaves pretty much everything as an exercise, which is a bit overwhelming to me. :(
Ok. I made rather the opposite experience, our first part of the differential geometry course was containing a lot of topology so I had to revise a lot.
I'm also trying to improve my (non-existent) knowledge of algebra: I'm about 80% through the book by Vinberg, but I think I need another book. However, reading books alone is quite boring and I find myself not really focusing on a lot of the material that seems not all-too-relevant for geometry and stuff :P ...but then I end up not knowing what one needs to in order to get started on the more interesting things, of course.
Lemma 1.
For any $x$ in the ring $F[t,t^{-1}]$ ($F[t,t^{-1}]$: the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$), $x$ is a power of $t$ if and only if $x$ divides $1$ and $t-1$ divides $x-1$ (the divisibilities are meant, of course, in $F[t, t^{-1}]$).
Lemma 2.
...
@Huy: I can provide intuition for some apparently nonintuitive stuff if people ask for it, but... can't really make someone focus on something if he doesn't feel like it :)
@SohamChowdhury: For example the stated motivation. "We want to be able to multiply vectors such that they behave bilinearly." This wouldn't make me care about this tensor product at all.
@SohamChowdhury: Also, generally, I find it meaningless to state "this concept is very important, and/because it has a lot of applications" whilst not giving any example.
They have all sorts of nice properties and you can reduce any question about stuff in general to how the function behaves on a basis. That's nice, IMO.
@SohamChowdhury: My point was that instead of studying bilinear maps, the tensor product allows you to study linear maps, which is a lot easier. However the motivation stated in the book is just "we want to multiply vectors" and "this product has a lot of important applications", which are rather empty statements, at least to me.
@SohamChowdhury: I was more criticizing style rather than content. For example, if I teach differentiation, I don't start with "in the following classes we will compute the slope of functions because they're useful".
Maybe I'm just very different, but I need proper motivation to get me interested in a new topic. :(
Take a topological space X. Take the (there is only one unique, yes) simply connected cover of X, Y. Let G_X be the group of all self-homeomorphisms of Y that descend to the same map in X.
"Im Juni letztes Jahr" oder "Im Juni letzten Jahres" oder "Im Juni vom letzten Jahr". The first one is what you'd use when speaking, the second is probably the most correct way written, the third I wouldn't advice to use.
@SohamChowdhury: Maybe easier: "letzten Juni", but people have different conventions what "last June" means, but to me it would mean Juni last year. :D
Find a topological space $X$ that has a dense, isolated (proper) subset $S\subset X$. (That is, every open set in $X$ intersects $S$, and every point in $s\in S$ has a neighborhood $N_s$ such that $N_s\cap X=\{s\}$.)
(By "isolated subset" I mean every point in it is isolated in $X$.)
Answer: $\{0\}\cup\{1,1/2,1/3,1/4,\dots\}$ with the usual topology. The $1/k$ elements are dense and isolated.
you can't tensor $\mathbb R^3$ over $\mathbb C$, because $\mathbb R^3$ is not a complex vector space. what you mean to say is that the result, considered as a complex vector space, is 3-dimensional.
again, don't say "tensoring over R" as if it was opposed to "tensoring over C". you cannot write $V \otimes_{\mathbb C} W$ unless $V$ and $W$ are both complex vector spaces.
I don't know why you referred to the topologist's sine curve. A much easier curve here would look like a sawtooth. B is not possible as described in one of the answers.
Some linear algebra I never learned: if $T_1^ 1(V)$ is the space of multilinear maps $V^* \times V \to \mathbb{R}$, I wish to find a basis-independent isomorphism $T_1^1(V) \cong \text{End}(V)$. So I guess the obvious map $\text{End}(V) \to T_1^1(V)$ takes $\varphi$ and sends it to the map $(\omega, v) \mapsto \omega(\varphi(v))$. I have some trouble getting the inverse without picking a basis, I suppose I could send the map $F \colon V^* \times V \to \mathbb{R}$ to the map $V \to V^*$ sending
$v \mapsto F(-, v)$. Or, I suppose that is to $V^{**}$, however it sort of feels like cheating given the requirement of naturality. I do lack experience with this stuff, so it could be that everything I'm saying now is silly.
I have thought about this in the past. I don't know how to write down the inverse without a basis. But I'm satisfied since my original map was defined basis-free.
Yes, its a bijective linear map, so there is an inverse. Whether or not it is the inverse is of less importance for theoretical concerns. Thanks for the help as always.
Broadly: Lie groups. The speakers were very different though, the one I liked the best did essentially representation theory with a differential goemetric feel to it.
I know I will continue doing math, however there is no pressure in Norway: for masters its not very competitive, and we do not have the integrated masters/phd programs as is common in the US.
Haha, yes. So we'll see, I might stay and try my luck as an exchange student for a semester.
The problem with staying at the same uni all the time is that you do not tend to get as wide a range of contacts which seems to be invaluable in academia.
I technically have to take more courses but that's not a particularly serious requirement. (I have to take four more courses, and I have... four years to do it.) I have an advisor and a project, so right now I'm just reading towards that end.
Ah, cool. I'm trying to go crazy on the courseload in this period: I know enough to take reasonably hard courses, not enough to seriously consider any interesting questions.
No, I start work as a TA tomorrow, then after that week there will be a Norwegian tradition of students getting awfully drunk, then the semester will start for real the week after that.
Commutative algebra, Complex Analysis, Algebraic Topology, Cohomology and Differential geometry. As for the second question: I have no idea, but likely something algebraic in spirit.
Although it is also common to take a lot of easy courses, as you have ridiculously much freedom with your degree. I could take seven courses in literature and still get a math degree while not doing more than 30ECTS a semester.
Yes, I suppose so, if they don't need things to be overly formal that is fine.
I think I'll ask about that identification on math.SE, it is the first exercise in my diff. geo book. I feel kind of screwed if I don't have a proper solution.
OK, so given a map $\varphi: V^* \times V \to \mathbb R$, I guess you're right, you get a map $V \to V^{\ast \ast}$. But there is a natural isomorphism $V \to V^{\ast \ast}$, so invert that.
The isomorphism is just $v \mapsto (w \mapsto w(v))$.
I don't think it matters that it would be hard to write down the inverse of this map. The inverse of a naturally defined map is still naturally defined, no?
This reminds me of something from earlier: just because you write down the map with respect to a basis doesn't mean it's not naturally defined. Stupid example: the map $e_i \mapsto e_i$ doesn't depend on the choice of basis.
(I think we're misusing naturally. Don't we just mean canonical here? i.e. defined without depending on a basis?)
So the inverse of the map that I defined above, $V \otimes V^* \to \text{End}(V)$, can be defined by picking a basis $e_i$ of $V$ and writing $\varphi \mapsto \sum e_i \otimes e_i^*(\varphi)$, where by that I mean the map $w \mapsto $ the coefficient of $e_i$ in $\varphi(w)$. On account of it's the inverse of a map that we defined without needing a basis, this doesn't depend on the basis.
Of course, this isn't quite your question, because "bilinear maps $V^* \times V \to \mathbb R$ is probably, like, $(V^* \otimes V)^*$ or something? I'm not sure. But I think you can massage it to working.
this is cute (taken from Grothendieck's manuscripts "Pursuing Stacks". it has been TeXed up recently : look in the homotopy theory chat, whoever wants a copy)
I understand the chain-it's the explicit iso. on the right Andrew wanted.
I can see how to do it with a basis for $V$: write $v \in V$ in that basis, write $f \in V^{\ast}$ in the dual basis, and use that to define the linear endomorphism $T$.
Lemma 1.
For any $x$ in the ring $F[t,t^{-1}]$ ($F[t,t^{-1}]$: the polynomials in $t$ and $t^{-1}$ with coefficients in the field $F$), $x$ is a power of $t$ if and only if $x$ divides $1$ and $t-1$ divides $x-1$ (the divisibilities are meant, of course, in $F[t, t^{-1}]$).
Lemma 2.
...
