most recent 30 from http://mathoverflow.net 2013-06-20T13:34:38Z http://mathoverflow.net/feeds/user/1537 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/14530/4-genus-of-a-2-bridge-link Megan 2010-02-07T18:51:43Z 2011-06-20T16:48:36Z

<p>How can we calculate the 4-genus of a link L? The 4-genus is defined to be the minimal genus of orientable surface bounded by L in B^4. Is there any routine method to calculate that?</p> <p>Especially, any good idea how to calculate the 4-genus of a 2-bridge link? Thanks.</p>

http://mathoverflow.net/questions/37690/isomorphism-between-two-universal-p-typical-formal-group-laws/44515#44515 Megan 2010-11-02T04:06:02Z 2010-11-02T04:06:02Z

<p>I happened to see your question. And I don't know whether you've found a good reference. You may want to have a look at Haynes Miller's notes on cobordism. Below is a link for that: <a href="http://www-math.mit.edu/~hrm/papers/cobordism.pdf" rel="nofollow">http://www-math.mit.edu/~hrm/papers/cobordism.pdf</a></p> <p>The answer to your question is YES. It's discussed in the 5th section of Chapter 2 in the note.</p>

http://mathoverflow.net/questions/16711/homotopy-associative-h-space-and-coh-space Megan 2010-02-28T17:57:29Z 2010-03-01T20:48:08Z

<p>Let $[X, Y]_0$ denote base point preserving homotopy classes of maps $X\rightarrow Y$. A multiplication on a pointed space $Y$ is a map $\phi: Y\times Y\rightarrow Y.$ From this map, we can define a continuous map for each pointed space $X$, $\phi_X: [X, Y]_0\times [X, Y]_0\rightarrow [X, Y]_0,$ by the composition $$\phi_X (\alpha, \beta)(x)=\phi(\alpha(x), \beta(x)).$$ If $([X, Y]_0, \phi_X)$ is a group for each $X$, then $(Y, \phi)$ is called a homotopy associative $H$-space.</p> <p>A $coH$-space is defined from a comultiplication, namely, a map $\psi: X\rightarrow X\vee X.$ Then, for each pointed space $Y$, we can define a function $\psi^Y: [X, Y]_0\times [X, Y]_0\rightarrow [X, Y]_0$ in this way: $$\psi^Y(\alpha, \beta)=(\alpha\vee\beta)\circ\psi.$$ If $([X, Y]_0, \psi^Y)$ is a group for each $Y$, then $(X, \psi)$ is called a homotopy associative $coH$-space.</p> <p>So, as we can see, if we have a homotopy associative $coH$-space $(X, \psi)$ and a homotopy associative $H$-space $(Y, \phi)$, then we can define two group structures on the space $[X, Y]_0$. My question is: are they "equivalent" in some sense? Obviously, whatever $\phi$ or $\psi$ is, the zero element of the group is the constant map in $[X, Y]_0.$ However, the two group structures do depend on the choice of $\phi$ and $\psi$, which seems have little relationship with each other.</p>

http://mathoverflow.net/questions/14505/if-the-4-genus-of-a-link-is-zero-is-it-a-slice-link Megan 2010-02-07T16:18:53Z 2010-02-07T20:30:53Z

<p>An n-component slice link is a link that bounds n disjoint discs in B^4. And the 4-genus of a link is defined to be the minimal genus of orientable surfaces bounded by it in B^4.</p> <p>My question is: if the link bounds a surface with zero genus in B^4, is it necessarily a slice link? If not, any counter examples?</p>

http://mathoverflow.net/questions/51239/representations-of-semidirect-products-of-symmetric-groups/51241#51241 Megan 2012-08-22T00:02:43Z 2012-08-22T00:02:43Z

Hey, I'm interested in similar questions. I've checked the book your recommended...Do you know any good reference on the representation ring structure of wreath products? Especially the representation theory for the product G\wr S_n where G is not finite. Thanks!

http://mathoverflow.net/questions/16711/homotopy-associative-h-space-and-coh-space/16713#16713 Megan 2010-02-28T21:52:34Z 2010-02-28T21:52:34Z

Since $[SX, Y]_0=[X, \Omega Y]_0,$ let $X$ be a sphere, we get the conclusion. Thanks, Chris!

http://mathoverflow.net/questions/16711/homotopy-associative-h-space-and-coh-space/16713#16713 Megan 2010-02-28T21:52:27Z 2010-02-28T21:52:27Z

I see how to prove the commutativity of $\pi_i$ with $i\geq 2$. For pointed spaces $X$ and $Y$, let's consider the loop space $\Omega Y$ and the suspension $SX$. Define $\phi: \Omega Y\times\Omega Y\rightarrow\Omega Y$ by $\phi(u,v)=u(2t)$ when $t\in [0,\frac{1}{2}]$ and $=v(2t-1)$ when $t\in [\frac{1}{2},1].$ Then, $(\Omega Y,\phi)$ is a $H$-space. And, define $\psi: SX\rightarrow SX\vee SX$ to be $\psi([x,t])=([x,2t],*)$ when $t\in [0,\frac{1}{2}]$ and $=(*, [x,2t-1])$ when $t\in [\frac{1}{2},1]$. Then, $(SX, \psi)$ is a $coH$-space.

http://mathoverflow.net/questions/16711/homotopy-associative-h-space-and-coh-space/16713#16713 Megan 2010-02-28T20:35:41Z 2010-02-28T20:35:41Z

Thanks! I really love the &quot;distributivity&quot; equality you gave. It's easy to prove, useful, and elegant. With this formula, we can also prove that both groups are abelian. Thanks!

http://mathoverflow.net/questions/14530/4-genus-of-a-2-bridge-link/14871#14871 Megan 2010-02-11T06:08:23Z 2010-02-11T06:08:23Z

Thanks. I'm not quite sure what you denote by $\tau$. Is it the Thurston-Bennequin invariant? I thought KR might be a better one since it's more sophisticated than many other invariants.But it may be a good news for me that KR doesn't give a better bound. I don't need to be involved in the tough computation of KR any more.

http://mathoverflow.net/questions/14530/4-genus-of-a-2-bridge-link/14871#14871 Megan 2010-02-10T18:08:26Z 2010-02-10T18:08:26Z

Aha! Thanks. I'm reading your paper &quot;a slice genus lower bound from $sl(n)$ KR homology&quot; these days.

http://mathoverflow.net/questions/14530/4-genus-of-a-2-bridge-link/14544#14544 Megan 2010-02-09T05:05:49Z 2010-02-09T05:05:49Z

Thanks. I think KR homology is more interesting because it can give a finer lower bound for 4-genus. As for Rasmussen's s-invariant, it equals the signature when the link is alternating, which seems kind of trivial.

http://mathoverflow.net/questions/14530/4-genus-of-a-2-bridge-link/14534#14534 Megan 2010-02-07T21:06:32Z 2010-02-07T21:06:32Z

Thanks a lot. They are really helpful but don't apply to the case I'm interested in. It seems that not the concordance order of ALL 2-bridges knots K(p,q) is determined but only for those with p odd. And the knot I'm studying has p=208, which is neither odd nor a square.

http://mathoverflow.net/questions/14505/if-the-4-genus-of-a-link-is-zero-is-it-a-slice-link Megan 2010-02-07T18:57:50Z 2010-02-07T18:57:50Z

Thanks. And I realize the initial question was a stupid one. I edited it and think this one is the one I really wanna ask.