Recent content by zhentil

(t+\epsilon/2,\infty) is an open set in R. Hence its pre-image is open by definition of continuity. (x,y) lies in that set.

This is one of those "apply the definitions" problems. Let x \in g^{-1}(t,\infty) . By definition, there exists \epsilon >0 such that g(x)>t+2\epsilon. By definition of supremum, there exists y \in Y such that f(x,y)>t+\epsilon. By definition of continuity, there exists an open set A...

This is not one of those cases. If Taylor's claim is correct, his terminology is non-standard.

But the second one is easy. If the polynomial has a root in your field, the Galois group is trivial. If it doesn't, the galois group is the only group with two elements.

Where do these polynomials live? You have to fix a base field in order to talk about Galois groups.

Lee's also wonderful about answering emails with questions about his books. I spotted an error in that book and sent him an email, and he had emailed me back and posted the erratum within 24 hours.

Positive definite is defined using an inner product, so the post above should be taken as a definition, with the proviso that <Mx,x> > 0 for all non-zero x. The usual way to prove such things is to characterize the eigenvalues. In infinite dimensions, things get quite complicated, so one...

Huh? Is the question about which word was being defined in the definition?

I'm a bit confused when you say that g is a function of z and not its conjugate, hence the partial must be zero. Doesn't that only work when the variables are independent?

The way you've written it, you've chosen one y from one set. In any event, you can't be invoking the axiom of choice if you can prove that your collection is a set without invoking it.

There are two points to clear up. The first is that the Euler class of a sphere bundle is just the Euler class of the corresponding vector bundle. The second is that the Euler class of the corresponding vector bundle is defined using the Thom isomorphism. For your other question, the Euler...

Steenrod's "The Topology of Fibre Bundles" is the old standby.

Using the Euclidean metric in this case (or any metric in general), one maps a form a to the unique vector field X satisfying a(v)=<X,v> for all tangent vectors v.

In this context, it only works if H is a deformation retract of G. For the "full power," I think one has to go with either Cech cohomology or classifying spaces. It remains true, however.

I'm not sure what you mean by fibre bundles in general here. For a general fibre bundle, the structure group is the diffeomorphism group of the fiber. This is an infinite dimensional Lie group, but not what you have in mind, I would guess. A bundle homotopy between two vector bundles E_0 and...