bio | website | math.binghamton.edu/somnath |
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location | Binghamton | |
age | 32 | |
visits | member for | 4 years, 8 months |
seen | Jul 12 at 11:00 | |
stats | profile views | 2,618 |
Passionate about mathematics.
Jul 2 |
awarded | Curious |
Nov 22 |
answered | Is there a good way to understand the free loop space of a sphere? |
Nov 21 |
awarded | Yearling |
Oct 10 |
revised |
Are Sasakian metrics (associated to bumpy metrics) bumpy?
added 1 characters in body |
Oct 10 |
asked | Are Sasakian metrics (associated to bumpy metrics) bumpy? |
Sep 6 |
revised |
Cobordism and finite sheeted covers of manifolds
question made more specific |
Sep 5 |
comment |
Cobordism and finite sheeted covers of manifolds
The fact that characteristic numbers (Pontrjagin and Stiefel-Whitney numbers) determine cobordism classes completely is true for compact, oriented manifolds. What is a reference for this being true for manifolds which are not necessarily compact? |
Sep 5 |
comment |
Cobordism and finite sheeted covers of manifolds
@ Sam - Thanks! I made the necessary changes. |
Sep 5 |
revised |
Cobordism and finite sheeted covers of manifolds
added 21 characters in body |
Sep 5 |
asked | Cobordism and finite sheeted covers of manifolds |
Aug 30 |
comment |
loop space homology and lens spaces
Do you mean for each component of the free loop space? The free loop space $LM$ has $p$ components for the lens space $M=L(p,q)$. |
Jun 25 |
awarded | Promoter |
May 13 |
awarded | Good Question |
Feb 19 |
comment |
Homology of classifying space of spin group BSpin(n)
@Xiao-Gang : Write the spectral sequence (in homology) for the fibration $G\to EG\to BG$. It will follow from the $E^5$ and $E^6$ page that $H_5(BSpin(n);\mathbb{Z})=H_4(Spin(n);\mathbb{Z})$ and $H_6(BSpin(n);\mathbb{Z})=H_5(Spin(n);\mathbb{Z})$. It follows from Milnor-Moore (with your input on the homotopy groups of $Spin(n)$) that these are torsion groups. Now you just have to compute these! |
Feb 19 |
comment |
Homology of classifying space of spin group BSpin(n)
@Xiao-Gang : Write the spectral sequence (in homology) for the fibration $G\to EG\to BG$. It will follow from the $E^5$ and $E^$ page that $H_5(BSpin(n);\mathbb{Z})=H_4(Spin(n);\Z)$ and $H_6(BSpin(n);\mathbb{Z})=H_5(Spin(n);\Z)$. It follows from Milnor-Moore (with your input on the homotopy groups of $Spin(n)$) that these are torsion groups. Now you just have to compute these! |
Dec 1 |
awarded | Popular Question |
Nov 30 |
answered | Homology of classifying space of spin group BSpin(n) |
Nov 21 |
awarded | Yearling |
Sep 20 |
comment |
Heegard Floer homology
This is a contentious, debatable comment that doesn't answer the question asked by OP. |
Sep 19 |
comment |
Whitehead product with identity on homotopy groups of spheres
Thanks for your answer! That's what (that $[x,x]=0$) I was thinking naively but I dug up some literature and realized that it's not the case. It's non-zero when $n$ is odd and generates a $\mathbb{Z}_2$. When $n$ is even it generates an infinite cyclic subgroup which splits on occasion. Judging from old papers, it seems that it would be too stupid (on my part) to expect a complete answer. Among other things, it would grossly underestimate the intricacies involved in knowing the homotopy groups of spheres! |