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Graduate student in mathematics.


Apr
24
revised Classification of general fibre bundles
added 21 characters in body
Apr
24
comment Classification of general fibre bundles
@Matt: in the rank $k$ vector bundle case the base space is the Grassmannian of $k$-planes in $\mathbb R^\infty$. Does this really generalize in the proof you saw to give the base space when the fiber is arbitrary?
Apr
24
comment Classification of general fibre bundles
I should have mentioned in my question that I know what I said holds true for $G$ a Lie group. But does this still hold if $G$ is any topological group? Also, if $F$ is a CW complex is $Aut(F)$ also?
Apr
24
asked Classification of general fibre bundles
Apr
21
comment Showing a hypersurface is contained in a level set of a regular value
Say $H(S) = c$. Then in general $H^{-1}(c)$ will have to contain more than $S$. How can I guarantee other points in $H^{-1}(c)$ are not critical?
Apr
21
comment Showing a hypersurface is contained in a level set of a regular value
@user8268: It looks like in your construction $S = H^{-1}(0)$? That can't work in general.
Apr
21
comment Showing a hypersurface is contained in a level set of a regular value
@user8268: Can you be more specific? If I simply extend the map by multiplying by a bump function then 0 will no longer be a regular value.
Apr
21
comment Showing a hypersurface is contained in a level set of a regular value
Do you really mean composition? $pr_2$ is not defined on all of $M$, only on an open set. If I multiply $pr_2$ by a bump function to extend it to $M$ then 0 will not be a regular value (since the function will be identically zero off of $N$).
Apr
21
answered What is the second stiefel whitney class of SO(n)?
Apr
21
asked Showing a hypersurface is contained in a level set of a regular value
Apr
13
comment What are the polar coordinates of the origin?
Sorry but how is this relevant to the question?
Apr
8
comment $K(\mathbb R P^n)$ from $K(\mathbb C P^k)$
@user7887: nevermind, we can just look at $2(k-1)x$.
Apr
8
comment $K(\mathbb R P^n)$ from $K(\mathbb C P^k)$
@user7887: i'm still confused, at the $k$th step we do not know if the group is $Z_2 \oplus Z_{2^{k-1}}$ or $Z_{2^k}$. Showing that there is an element $x$ such that $2x$ is not trivial does not distinguish these in general (only for the $RP^4$ case).
Apr
7
comment Suppose that $x > 1$. Prove that $s_n \to 1$
Showing 1 is a lower bound then means 0 can't be the limit. But just showing that 0 is a lower bound does not rule out 0 as the limit (so your proof is not complete).
Apr
7
comment Suppose that $x > 1$. Prove that $s_n \to 1$
I don't see how you can conclude that the limit must be 1. You know its decreasing and that the limit is either 0 or 1. It starts above 1 so why can't it decrease all the way to 0? However, it's not too hard to show that 1 is a lower bound as well.
Apr
7
comment $K(\mathbb R P^n)$ from $K(\mathbb C P^k)$
@user7887: I spent some time on it but was not able to get anywhere. I was also unable to find what you referenced in my edition of Atiyah.
Apr
6
comment $K(\mathbb R P^n)$ from $K(\mathbb C P^k)$
I don't understand the Stiefel-Whitney class argument. First of all $w(RP^n)$ vanishes if $n+1$ is a power of two. Secondly we are considering complex vector bundles and it is possible for the complexification of a non-trivial real vector bundle to be trivial (e.g. the complexification of the mobius bundle).
Apr
2
comment $K(\mathbb R P^n)$ from $K(\mathbb C P^k)$
Thanks for the answer. I'm not familiar with the Veronese embedding. I found this en.wikipedia.org/wiki/Veronese_surface#Veronese_map but I don't think this is what you're talking about since this map doesn't always give a map $P^n \to P^{2n+1}$. Also, how do you see from the spectral sequence that the map is surjective?
Apr
1
awarded  Nice Question
Mar
30
revised Is correct to say that every tensor is a spinor but not every spinor is a tensor?
edited body