transition matrices? I am thinking of transition functions as those things from differential geometry, what is a transition matrix in the context of line bundles?
The house I just sold for about $225K in Georgia (with 1 acre + creek) would go for well over a million in places I'd want to live in CA, probably more than that.
@AlexWertheim $L/k$ be a separable field extension. Pushout (as $k$-modules) of the diagram $L \hookleftarrow k \hookrightarrow k^{alg}$ is $L \otimes_k k^{alg}$. Now, the map $L \to L \otimes_k k^{alg}$ should somehow relate to $\mathsf{Hom}_k(L, k^{alg})$ (well, $L \otimes k^{alg}$ is isom to a direct sum of copies of $k^{alg}$, and maps $\prod k^{alg} \to L$ should be analogous to the map $\bigsqcup * \to X$, as products translates to coproducts in the algebraic and topological picture).
@AlexWertheim I am trying to find a way to get a formal analogy between covering spaces and galois theory. so far, I have found this : embeddings $k \hookrightarrow k^{alg}$ are analogous to point, morphisms between two embeddings (an aut of k^alg s.t. the diagram commutes) is analogous to a path, a neighborhood of a point is a $k$-algebra over $k$ s.t. a diagram commutes, again. fibers is either $A \otimes_k k^{alg}$ or $\mathsf{Hom}_k(L, k^{alg})$. ideally, it should be both.
@MikeM: that sounds about right. Actually, embarrassing as this is, I'm dreading the basic qual. How effective is boot camp for the analysis side? (I suppose this varies from person to person)
@TedShifrin My thought process was to find the combination of getting two reds and one blue and multiplying by 3! because that is how many ways you can arrange it.
Here was the first question on my probability final last fall, @Deathslice: A committee of $7$ people is to be chosen from a group consisting of $9$ (distinguishable) women and $10$ (distinguishable) men. If the committee must contain at least $3$ women and at least $2$ men, how many different committees are possible?
You can do the same approach for your next question, but you have to make sure you account for exactly one $1$ in your combinations count.
Is an injective map $f : \mathbb{R}^2 \to \mathbb{R}^2$ that sends lines to lines necessarily a linear transformation? I'm not totally sure, but I can't think of a good counterexample.
so I considered the set $(-\infty,-1] \cup [1,\infty)$ and constructed the ball by considering if it is in the left set the one that has -1 then I consider y + 1 / 2 and the other one is y - 1 / 2
now the problem I am having for points inside of the interval (-1,11)
Your set is a union of two sequences, $\{-1 + 1/n\}$ and $\{1 + 1/n\}$. You should prove that for any point other than 1, -1 in the sequence, there is a ball disjoint from any other point. This proves that no such point can be an accumulation point.
yeah that is what I am doing @BalarkaSen I am proving that there -1 and 1 are accumulation points and there doesn't exists any other accumulation points
But you should be able to argue that for any point other than $\pm 1$ any interval *not containing $\pm 1$ contains at most finitely many points of the set.
But there's a much better way to do this. If you pick any point $y$ other than $\pm 1$, you should be able to describe how to pick an interval that contains no point of the set, other than possibly $y$ itself.
I'm wondering how should I spend the next 30 minutes I have, relearning tensor products, or racking my brains over on the problem I am thinking about, or hunting for an algebraist to help with my work, or just sleeping.
of course, doing any of these have advantages over the others. but the optimal advantageous strategy is what I want.
@TedShifrin, Thanks for the help, you were right. If $f$ is bijective, continuous, preserves lines, and fixes the origin then it's a linear transformation.
