most recent 30 from http://mathoverflow.net 2013-05-22T07:15:54Z http://mathoverflow.net/feeds/user/1182 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20877/what-is-the-difference-between-homology-and-cohomology/20885#20885 Brad Hannigan-Daley 2010-04-10T02:17:52Z 2013-03-12T04:20:22Z

<p>As to what cohomology actually measures, I think a general theme is "the failure of locally trivial things to be globally trivial", or perhaps "the failure of local solutions to glue together to form a global solution". In the de Rham cohomology of a smooth manifold, any closed form &omega; is "locally trivial" in that you can cover the manifold by contractible charts, over each of which a solution to d&alpha;=&omega; exists by the Poincaré lemma. The cohomology class [&omega;] measures the failure of existence of a global solution of this equation. Similar remarks can be made about simplicial, singular, and (especially) Čech cohomology.</p>

http://mathoverflow.net/questions/108734/theorem-on-composition-of-derived-functors-question-about-proof/108736#108736 Brad Hannigan-Daley 2012-10-03T18:47:09Z 2012-10-04T21:11:19Z

<p>To show that $E$ is an isomorphism of functors, it suffices to show that $E(A)$ is an isomorphism for each object $A$ of $D^+(\mathcal{A})$. This has been shown for each $K^\bullet$ an object of $\operatorname{Kom}^+(\mathcal{R}_\mathcal{A})$. For an arbitrary object $A$, choose a quasi-isomorphism $f:A\to K^\bullet$ for such a $K^\bullet$, which then becomes an isomorphism in the derived category. Then $R(G\circ F)(f):R(G\circ F)(K)\to R(G\circ F)(A)$ and $(RG\circ RF)(f):(RG\circ RF)(K^\bullet)\to(RG\circ RF)(A)$ are isomorphisms. We then have that $E(A) = (RG\circ RF)(f)\circ E(K^\bullet) \circ R(G\circ F)(f)^{-1}$ is an isomorphism, being a composition of isomorphisms.</p>

http://mathoverflow.net/questions/33193/reduced-rings-idempotents-and-their-prime-spectrum/33200#33200 Brad Hannigan-Daley 2010-07-24T14:06:18Z 2010-07-24T15:04:17Z

<p>In general, if $f:R\to S$ is a surjective ring homomorphism, then the map Spec $S\to$ Spec $R$ is a closed embedding (in particular, it's injective).</p> <p>Edit: I missed that $e$ may not necessarily be in $A$; the following assumes that it is.</p> <p>If $e$ is an idempotent then so is $1-e$, and $A$ decomposes as a product $A = Ae \times A(1-e)$. From the perspective of schemes, this product turns into a coproduct, and Spec $A$ is the disjoint union of Spec $Ae$ and Spec $A(1-e)$. The schemes of the form Spec $Ae$ are hence those obtained as a union of some subset of the connected components of Spec $A$.</p>

http://mathoverflow.net/questions/8428/freeing-a-sphere-from-within-a-sphere Brad Hannigan-Daley 2009-12-10T07:03:40Z 2010-07-07T01:20:06Z

<p>We can embed $S^2\times I$ into $\mathbb{R}^3$ by taking a compact 3-ball and removing an open 3-ball from its interior. Taking the boundary gives an embedding $i: S^2\sqcup S^2\hookrightarrow\mathbb{R}^3$, as "a sphere contained inside another sphere". Now it's intuitively clear that this embedding is not ambient-isotopic to the embedding $j$ given by putting these two spheres "side-by-side" in $\mathbb{R}^3$. That is, there is no isotopy $F_t:\mathbb{R}^3\to \mathbb{R}^3$ with $F_0=1_{\mathbb{R}^3}$ and $F_1\circ i=j$. At least, this looks visually obvious. Assuming my intuition isn't betraying me and this isn't false, what's an elegant way to prove this? </p> <p>Also, (why) does there (not) exist an embedding $S^2\times I\hookrightarrow\mathbb{R}^3$ whose boundary <em>is</em> ambient-isotopic to the "side-by-side" embedding? How about for $S^{n-1}\times I\hookrightarrow\mathbb{R}^n$?</p>

http://mathoverflow.net/questions/21995/name-for-topology-making-group-action-continuous/22006#22006 Brad Hannigan-Daley 2010-04-21T02:37:14Z 2010-04-21T03:15:51Z

<p>If the action of $G$ on $X$ is continuous (i.e. the multiplication map $X\times G\to X$ is continuous) then the resulting topology <em>is</em> $\tau$:</p> <p>Let $\tilde X$ denote $(X\times G)/\sim$, and let $\phi:X\to \tilde X:x\mapsto[x,e]$ be the identification you mentioned (with the factors $X,G$ reversed for convenience). Then $\phi$ is clearly continuous. Let $\psi:\tilde X\to X$ be the inverse of $\phi$, i.e. $\psi([x,g]) = xg$. For an open subset $U$ of $X$, $\psi^{-1}(U) = {(x,g):xg\in U}$. Pulling this back to $X\times G$ via the projection $X\times G\to\tilde X$ gives exactly the preimage of $U$ under the multiplication map $X\times G\to G$, which is open, and so $\psi^{-1}(U)$ is open in $\tilde X$. So $\phi$ is a homeomorphism.</p>

http://mathoverflow.net/questions/1722/free-high-quality-mathematical-writing-online/8824#8824 Brad Hannigan-Daley 2009-12-14T02:12:25Z 2009-12-14T02:12:25Z

<p><a href="http://www.tac.mta.ca/tac/reprints/articles/17/tr17abs.html" rel="nofollow">Abstract and concrete categories: The joy of cats</a> by Jiri Adamek, Horst Herrlich and George Strecker, is a nice book for learning category theory. It went out of print, so the authors made it available online for free.</p>

http://mathoverflow.net/questions/5450/cocktail-party-math/5599#5599 Brad Hannigan-Daley 2009-11-15T02:24:43Z 2009-11-15T02:24:43Z

<p>I like to talk about planarity of graphs, because it's easy to introduce, easy to draw on a napkin, and on the "application" side of things you can mention computer chip design, transportation networks, and other things. The nonplanarity of $K_{3,3}$ can be introduced via the "three utilities" puzzle: you have three houses and you want to connect each one to each of three utilities (water, hydro, gas) -- can you do it without having the lines cross? Similarly the nonplanarity of $K_5$ can be posed as a puzzle, and Kuratowski's theorem basically asserts that these two examples give the only obstructions to planarity: a graph is planar if and only if it contains no subgraph homeomorphic to $K_5$ or $K_{3,3}$.</p> <p>Going further, you can talk about embedding graphs on a torus (i.e. in a game of Asteroids), for which you can draw examples of embeddings of $K_5$ and $K_{3,3}$, and say that there's an analogy of Kuratowski's theorem (by corollary to the Robertson-Seymour theorem): there is a finite list of "forbidden graphs" which are the only obstruction to a graph being embeddable on the torus, in the sense that any nonembeddable graph contains one of those forbidden graphs as a minor.</p>

http://mathoverflow.net/questions/3242/canonical-examples-of-algebraic-structures/3248#3248 Brad Hannigan-Daley 2009-10-29T12:06:46Z 2009-10-30T01:33:05Z

<p>This is, at least, how it went for me just now:</p> <pre><code>Read: Think: field Q(sqrt 2), or some other number field finite group symmetric group Lie group the torus R^2/Z^2 ring a matrix ring over... some other ring commutative ring a polynomial ring k[x_1,...,x_n] Lie algebra sl(2) </code></pre>

http://mathoverflow.net/questions/761/undergraduate-level-math-books/2910#2910 Brad Hannigan-Daley 2009-10-27T22:11:28Z 2009-10-27T22:11:28Z

<p>Dummit and Foote's <em>Abstract Algebra</em> is an excellent book for learning group theory, ring theory, and module theory. There's also a section on basic algebraic geometry and homological algebra.</p>

http://mathoverflow.net/questions/128891/what-does-the-q-catalan-numbers-count Brad Hannigan-Daley 2013-04-27T12:21:18Z 2013-04-27T12:21:18Z

@Adrien Hardy: They might still have enumerative significance. For example, the $q$-binomial coefficient ${n\brack k}_q$, when $q$ is specialized to a prime power, counts the number of $k$-dimensional subspaces of $(\mathbb{F}_q)^n$.

http://mathoverflow.net/questions/108734/theorem-on-composition-of-derived-functors-question-about-proof/108736#108736 Brad Hannigan-Daley 2012-10-04T21:10:30Z 2012-10-04T21:10:30Z

I meant an isomorphism in the derived category. I'll edit it to make it clearer.

http://mathoverflow.net/questions/81715/my-way-of-realizing-the-relation-between-continuity-and-integrability Brad Hannigan-Daley 2011-11-23T15:05:30Z 2011-11-23T15:05:30Z

MathOverflow is for research-level questions. This question would be better suited to math.stackexchange.com.

http://mathoverflow.net/questions/81456/possible-borel-subgroups-of-gl-n/81458#81458 Brad Hannigan-Daley 2011-11-21T00:52:17Z 2011-11-21T00:52:17Z

When you say &quot;$\Phi$ must be closed under addition&quot; do you mean instead that the sum of two positive roots is either a positive root or not a root?

http://mathoverflow.net/questions/78830/finding-information-about-the-basics-of-an-advanced-math-topic Brad Hannigan-Daley 2011-10-22T17:02:24Z 2011-10-22T17:02:24Z

It's often worth searching in the arxiv to see if someone has written a relevant expository article. You could also try starting up a learning seminar and convincing other people to give talks in it :)

http://mathoverflow.net/questions/61921/on-closed-totally-disconnected-subgroups-of-connected-real-lie-groups/61923#61923 Brad Hannigan-Daley 2011-04-16T16:42:24Z 2011-04-16T16:42:24Z

Cartan's theorem implies that $H$ is necessarily a smooth submanifold of $G$. So if it's totally disconnected, it must be 0-dimensional, i.e. discrete.

http://mathoverflow.net/questions/56938/what-does-the-adjective-natural-actually-mean/56943#56943 Brad Hannigan-Daley 2011-03-01T00:56:21Z 2011-03-01T00:56:21Z

It should say &quot;the isomorphism $V\cong V^{**}$ between a <i>finite-dimensional</i> vector space and its double dual&quot; (given by $v\mapsto$ evaluation on $v$).

http://mathoverflow.net/questions/50573/some-questions-about-geometric-representation-theory Brad Hannigan-Daley 2010-12-28T20:42:52Z 2010-12-28T20:42:52Z

Please explicitly describe the action in &quot;Young Tableaux&quot; that you refer to -- this is more polite than expecting the reader to hunt down that book and find the exercise by themself in order to answer your question.

http://mathoverflow.net/questions/36309/tic-tac-toe-n-dimensional/36310#36310 Brad Hannigan-Daley 2010-08-21T21:54:45Z 2010-08-21T21:54:45Z

The way this is written makes it seem like you're suggesting the answer is 3^n-1 choose 2. (If this isn't the case, please ignore this comment!) But this is much bigger than the correct answer. The reasons for this are that (1) not every choice of two endpoints gives a line <i>through the centre</i>, and (2) for each line, there are several choices of a pair of endpoints which yield that line.

http://mathoverflow.net/questions/20664/why-is-complex-projective-space-triangulable/21790#21790 Brad Hannigan-Daley 2010-04-18T23:53:11Z 2010-04-18T23:53:11Z

Please stop doing whatever it is you are doing.

http://mathoverflow.net/questions/8428/freeing-a-sphere-from-within-a-sphere/8430#8430 Brad Hannigan-Daley 2009-12-10T07:38:32Z 2009-12-10T07:38:32Z

<i>slaps forehead</i> Thanks, can't believe I didn't notice that. I'm still curious about the (non-)existence of an embedding with that described property.