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Graduate student in mathematics.


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
Mar
30
answered Is correct to say that every tensor is a spinor but not every spinor is a tensor?
Mar
30
answered Exercise about smooth maps in Lee
Mar
30
comment Exercise about smooth maps in Lee
But you're working locally so take your coordinate neighborhood. Then take say the compact ball. Then you have a bump function which is identically 1 on the compact ball. But now look at the interior of the ball which is also homeomorphic to $\mathbb R^n$ and so gives you another chart in your atlas. So you can always take a chart contained in a compact set and then extend any function on that chart using a bump function.
Mar
30
comment Exercise about smooth maps in Lee
By using a bump function, you can extend any smooth function defined on only a subset to a global smooth function. I think this may help.
Mar
30
comment Exercise about smooth maps in Lee
I'm confused, $C^\infty(F^{-1}(V))$ are functions on a subset of $M$, so you can't pull these back via $F$. Maybe you want to show $C^\infty(F^{-1}(V)) \subset F^* C^\infty(V)$?
Mar
30
revised $K(\mathbb R P^n)$ from $K(\mathbb C P^k)$
added 956 characters in body
Mar
29
comment Unknown method (for me) to obtain the value of a variable
This is the "Fourier trick."
Mar
28
answered Lie algebra of a quotient of Lie groups
Mar
27
revised Why is the orthogonal group $\operatorname{O}(2n,\mathbb R)$ not the direct product of $\operatorname{SO}(2n, \mathbb R)$ and $\mathbb Z_2$?
added 98 characters in body
Mar
27
answered Why is the orthogonal group $\operatorname{O}(2n,\mathbb R)$ not the direct product of $\operatorname{SO}(2n, \mathbb R)$ and $\mathbb Z_2$?
Mar
26
comment mathematical existence
"to be is to be the value of a variable"
Mar
25
awarded  Taxonomist
Mar
23
comment Which is the “proper” definition of a geodesic curve?
I don't think there can ever be a longest path: if $\gamma$ is a path then we can compose $\gamma$ with the path taking us backwards along $\gamma$ and then compose again with $\gamma$ to get a path 3 times as long as $\gamma$. Edit: I'm guessing though you meant local maximum.
Mar
15
answered Quotient map in polynomial ring
Mar
14
comment $K(\mathbb R P^n)$ from $K(\mathbb C P^k)$
@user7887 $\mathbb R P^n$ is a quotient of $S^n$, not a subspace. The long exact sequence is for a pair $(X,Y)$ where $Y \subset X$.
Mar
13
comment vector space of all smooth functions has infinite dimension
You can use this fact to prove the desired result. Define a polynomial locally in a chart and extend it to the entire manifold by multiplying with a bump function. Then you'll get an infinite dimensional subspace of $C^\infty$ in this way.