1,598 reputation
11020
bio website math.binghamton.edu/somnath
location Binghamton
age 32
visits member for 4 years, 8 months
seen Jul 12 at 11:00
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!