tag:blogger.com,1999:blog-6128955598718204808.comments2016-06-02T10:07:15.096-07:00The Matthew Maths ShowMatthew Eric Bassetthttp://www.blogger.com/profile/10576806213820001905noreply@blogger.comBlogger21125tag:blogger.com,1999:blog-6128955598718204808.post-5405976654002920132016-06-02T10:07:15.096-07:002016-06-02T10:07:15.096-07:00You can ask here for a help. They helped me a lot ...You can ask here for a help. They helped me a lot an i`m highly satisfied with quality of work done. I can promise you 100% un-plagiarized text and good experts there. Use with pleasure! <a href="http://bit.ly/1X45IyA" rel="nofollow">Evolution Writers</a> Annahttp://www.blogger.com/profile/10910310615989848546noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-14453862144464321862013-03-08T08:45:12.050-08:002013-03-08T08:45:12.050-08:00I'm glad you found it helpful! It's been ...I'm glad you found it helpful! It's been so long ago I forgot exactly how I came up with it, but chances are that I got it from someone else, I should have linked to the referenced physics forum discussion. In any case, I fixed the $\LaTeX$ error. :)Matthew Eric Bassetthttp://www.blogger.com/profile/10576806213820001905noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-54485884944792365322013-03-07T21:44:36.846-08:002013-03-07T21:44:36.846-08:00Awesome proof , I tried to do this for about 2 hrs...Awesome proof , I tried to do this for about 2 hrs but could not get it , your proof really helped me. Thanks. btw could you try to correct that typo in "Extra close brace or missing open brace"? that would be nice , then people won't have to check the math for understanding it.Abhinav Tamaskarhttp://www.blogger.com/profile/09991413433109188398noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-3914720442767196792012-04-18T06:46:58.687-07:002012-04-18T06:46:58.687-07:00I was just thinking how as we we get older and hen...I was just thinking how as we we get older and hence wiser in our own view that we always think that we know everything (the perennial know-it-all attitude that has steathily crept up on me) but along the way we have forgotten about the basics which really are the foundation...and everyone knows the story of the house that was built without a foundation...Jerukimhttp://www.blogger.com/profile/04363911004855222490noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-28751219662352659592011-07-07T03:12:59.920-07:002011-07-07T03:12:59.920-07:00different perspective: why do you do this? sounds ...different perspective: why do you do this? sounds to me like you get some sort of pleasure to proof things to yourself (just because you can) and some pain (i could've/should've done this faster) afterwards. sounds like a wired mix of emotions that you want to generate for yourself for some reason. i don't know how that feels to you, but maybe it's worth a deeper exploration.mowsenhttp://www.blogger.com/profile/00428925771907982233noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-901322511384993722011-06-21T09:51:23.057-07:002011-06-21T09:51:23.057-07:00Once something clicks it is so simple. Try explain...Once something clicks it is so simple. Try explaining E=MC^2 to the average person, it's pretty difficult. Once you have used your imagination to properly extract the knowledge you have attained from any subject, things become much clearer and much more discover-able especially flaws in a system.Ron Zeligzonhttp://www.blogger.com/profile/11331864418126760739noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-8962611076908937582011-06-21T05:24:46.355-07:002011-06-21T05:24:46.355-07:00Schoppenhauer:
Every truth passes through three s...Schoppenhauer:<br /><br />Every truth passes through three stages before it is recognized In the first it is ridiculed, in the second it is opposed, in the third it is regarded as self-evident.Jörghttp://www.blogger.com/profile/10141628694582663144noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-30051065269591973592011-06-21T03:16:55.159-07:002011-06-21T03:16:55.159-07:00This comment has been removed by a blog administrator.henryhttp://www.blogger.com/profile/08180035343726711042noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-5289838440137565732011-06-20T19:35:47.888-07:002011-06-20T19:35:47.888-07:00A quote from Feynman seems appropriate,
"Wh...A quote from Feynman seems appropriate, <br /><br />"When you are thinking about something that you don’t understand you have a terrible, uncomfortable feeling called ‘confusion’. It’s a very difficult and unhappy business. So, most of the time you are rather unhappy, actually, with this confusion. You can’t penetrate this thing. Now, is the confusion… is it because we are all some kind of apes that are kind of stupid working against this? Trying to figure out to put the two sticks together to reach the banana and we can’t quite make it? …the idea ? And I get that feeling all the time: that I am an ape trying to put two sticks together. So I always feel stupid. Once in a while, though, everything — the sticks — go together on me and I reach the banana."drhodeshttp://www.blogger.com/profile/10482944964078869642noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-88342684385873498262010-11-15T07:26:23.529-08:002010-11-15T07:26:23.529-08:00Yes, the Haar measure thingy can be avoided in the...Yes, the Haar measure thingy can be avoided in the etale case. Being etale implies that the fibers of the source and target maps are discrete, but not necessarily finite. If your ambient manifold is compact, every discrete set is finite, so the sum is well defined for all functions. If the manifold is not compact the fibers may not be finite (think of the covering of the circle by real numbers, that is an etale map but fibers are infinite); this gets sorted out by picking functions with finite or compact support. In the etale case when one has compact support, the intersection of the fiber of the source or target map (which is discrete) with the support of the function (which is compact) must be both compact and discrete, thus finite and the sum is well defined.<br /><br />Away from the etale case I think there is no way to avoid the use of the Haar measure.Javier Lopezhttp://www.blogger.com/profile/01409120125923071820noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-15269268252539820782010-11-15T06:28:27.824-08:002010-11-15T06:28:27.824-08:00Yes and no.
In fact, it's a question I have (...Yes and no.<br /><br />In fact, it's a question I have (but didn't state.) I guess you have it too. I would think that in order to extend the def to a topological group[oid] you would need to fix a haar measure and define it convolution over the integral of the whole group. But neither Connes nor Khalkahali used such a construction for an Etale groupoid. I recall Khalkahali saying explicitly that you can avoid it for an Etale groupoid, and I am wondering why myself.<br /><br />I'm working on application essays atm, but I hope to drop by soon!Matthew Eric Bassetthttp://www.blogger.com/profile/10576806213820001905noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-68608358534544919122010-11-15T06:24:15.853-08:002010-11-15T06:24:15.853-08:00Are you sure about the extension of the definition...Are you sure about the extension of the definition for topological groupoids? If I remember correctly when defining the group algebra of a topological group one needs to fix a Haar measure in the group and replace the summation by integration against that measure. Same thing should be true for groupoids. The reason for choosing continuous functions is to keep track of the group(oid) topology, and the compact support to ensure the integrals are finite.<br /><br />Btw, I am back in London, in case you want to drop by my office.Javier Lopezhttp://www.blogger.com/profile/01409120125923071820noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-19104916436058932522010-10-25T03:22:06.107-07:002010-10-25T03:22:06.107-07:00Glad I could help (sorry it didn't arrive on t...Glad I could help (sorry it didn't arrive on time for your talk). I personally find easier to think of groupoids as certain kinds of directed graphs and work with a few finite examples to get some intuition. Infinite ones, though more geometrical, make the algebraic properties harder to grasp.Javier Lopezhttp://www.blogger.com/profile/01409120125923071820noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-45652872722801064752010-10-21T02:21:49.441-07:002010-10-21T02:21:49.441-07:00thanks for the comments!
"You can easily pro...thanks for the comments!<br /><br />"You can easily prove this identifying each function $f$ on $G$ with the element $sum_{x \in G} f(x)$ in $\mathbb{C}[G]$. Now one can prove that the group algebra is commutative if and only if $G$ is commutative."<br /><br />That's actually the most helpful statement. I was thinking yesterday of the equiv statement for a group algebra, if $G$ is a finite group, then $\mathbb{C}[G]$ is just the $\mathbb{C}$ algebra with $G$ as a basis and the group operation as multiplication. In that case it's obviously commutative if and only if G is commutative. But I wasn't able to see that in the function description of a groupoid. http://en.wikipedia.org/wiki/Convolution#Properties especially confused me. :)Matthew Eric Bassetthttp://www.blogger.com/profile/10576806213820001905noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-89552415391529252282010-10-20T01:28:49.157-07:002010-10-20T01:28:49.157-07:00An easy example that should convince you: take a s...An easy example that should convince you: take a set $X=\{a,b\}$ with two elements, consider the total equivalence relation $a\sim b$, and construct the convolution algebra for the equivalence groupoid (without bothering about completion). What algebra do you get?Javier Lopezhttp://www.blogger.com/profile/01409120125923071820noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-70416712203967102512010-10-20T01:24:45.030-07:002010-10-20T01:24:45.030-07:00No, they aren't! Think of the most simple exam...No, they aren't! Think of the most simple example of groupoid: a group $G$. You can look at it as the groupoid with only one object and arrows indexed by the elements of the group, i.e. $\mathcal{G}_0 = {\ast\}$ and $\mathcal{G}_1 = G$.<br /><br />The convolution algebra (forget about topology for now, even assume G is finite so no worries about infinite sums) is the set of functions from the arrows of the groupoid to the complex numbers. In this case it would be the set $\{ f:G \to \mathbb{C} \} $; now, in this set we can give two different algebra structures. If we take pointwise sum and multiplication we just obtain the commutative and boring $G^\mathbb{C}$, direct product of $|G|$ copies of the complex numbers. But if we take the convolution product instead, what one gets is the group-ring $\mathbb{C}[G]$. You can easily prove this identifying each function $f$ on $G$ with the element $\sum_{x\in G} f(x)\cdot x$ in $\mathbb{C}[G]$. Now one can prove that the group algebra is commutative if and only if $G$ is commutative.<br /><br />For general groupoids you have something similar. SInce every going arrow has the corresponding inverse coming back, one ends up having to deal with all the compositions of a given arrow with things in the isotropy group of the source or target. If the isotropy group is noncommutative (as it is for the BC-system) then the convolution algebra will also be noncommutative. Even if the isotropy group is commutative one often has to deal with a crossed-product action that becomes noncommutative.<br /><br />Dude, if things were just commutative we'd just use prime/primitive spectrums and work happily with classical geometry rather than undertaking all this major mess! ;-)Javier Lopezhttp://www.blogger.com/profile/01409120125923071820noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-77151029611617942932010-08-12T06:47:23.382-07:002010-08-12T06:47:23.382-07:00I'm reading two main monographs [plus associat...I'm reading two main monographs [plus associated papers]. Connes & Marcolli's Noncommutative Geometry, Quantum Fields, and Motives. And Marcolli's Lectures on Arithmetic Noncommutative Geometry. <br /><br />There's not really any textbooks for the main subject. But I have been able to chase down textbooks and monographs for tangents. For instance, R. Haag Local Quantum PHysics and Bratteli & Robinson Operator Algebras and Quantum Stastical Mechanics contains a lot of the QSM stuff, especially on those KMS states. That said both of those books are pretty far above my physics background and reference a lot of statistical thermo that I've never seen. I think it would be worth going through some course notes on a lower level course, but I just don't have the time.Matthew Eric Bassetthttp://www.blogger.com/profile/10576806213820001905noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-71531716603934785142010-08-11T06:09:23.459-07:002010-08-11T06:09:23.459-07:00In your project, are you pretty much using researc...In your project, are you pretty much using research papers or is there any books which you can read?<br /> The thing I find sometimes is that even when some result should be "clear" from the definition, it is hard to get a good understanding of the meaning of the result without a good intuition for what is going on. I guess it must help that this things relate to statistical thermodynamics/mechanics. It might even be worth reading some course notes in the subject to help in the intuition front.acyrlhttp://acyrl.wordpress.com/noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-32846012542410725852010-06-15T05:03:50.011-07:002010-06-15T05:03:50.011-07:00$...$ or $\ldots$ or Micro$oft$...$ or $\ldots$ or Micro$oftniko.laaksonenhttp://www.blogger.com/profile/07299107289011916185noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-39625287751513845332010-06-15T05:00:31.673-07:002010-06-15T05:00:31.673-07:00yes I can. just inside $ $. but what if I do som...yes I can. just inside $ $. but what if I do something like<br /><br />the prices range from $100 to $300.M.E. Bassetthttp://www.blogger.com/profile/10576806213820001905noreply@blogger.comtag:blogger.com,1999:blog-6128955598718204808.post-81532061706522991702010-06-15T03:36:32.189-07:002010-06-15T03:36:32.189-07:00can I use LaTeX too? $e^{x^2 + 5y} = \int^{\infty...can I use LaTeX too? $e^{x^2 + 5y} = \int^{\infty}_{0} f(x) dx$M.E. Bassetthttp://www.blogger.com/profile/10576806213820001905noreply@blogger.com