User anon - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T08:04:27Z http://mathoverflow.net/feeds/user/9429 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/48692/organizing-principles-of-mathematics/48715#48715 Answer by anon for Organizing principles of mathematics anon 2010-12-09T05:18:17Z 2010-12-09T05:18:17Z <p>The Choquet theory in convex analysis / functional analysis / whatever you want to call it. An element of a convex set should be some kind of "average" of extreme points. This has the status of a theorem for compact sets in normed linear spaces but is a useful guiding principle for not-necessarily-compact sets in not-necessarily-normed linear spaces. Chapter 14 in Lax's Functional Analysis book gives good examples of the wide array of applications of the same simple idea.</p> http://mathoverflow.net/questions/47850/associativity-of-polar-decomposition/47887#47887 Answer by anon for Associativity of polar decomposition anon 2010-12-01T10:28:49Z 2010-12-01T10:28:49Z <p>I have not fully checked this idea, but here goes. I prefer the notation $P(T)$ for the partial isometry occurring in the polar decomposition of $T$. I also got lost with three Hilbert spaces in the mix, so this answer is only for the case where $A,B,C$ all operate on the same fixed Hilbert space $H$.</p> <p>In this notation, I believe the question is whether or not<br> $$P(P(AB)C) = P(A P(BC))$$ holds for all partial isometries $A,B,C$ on $H$.</p> <p>I think it is easy to see that if $U$ is unitary, then for all $X$ we have $P(UX) = U P(X)$, and that if $T$ is a partial isometry, then $P(T) = T$. From this, it seems to follow that if $A$ is assumed unitary, and $B$ and $C$ are partial isometries, we have $P(P(AB)C) = P(AP(B)C) = P(ABC) = A P(BC)$ and $P(A P(BC)) = A P(P(BC)) = A P(BC)$ so the desired result holds.</p> <p>If $A$ is not unitary, it still has a unitary dilation. This means there is another Hilbert space $K$ and operators $X: K \to H$ and $Y: K \to K$ with the property that the operator $A'$ on $H \oplus K$ given by the block operator matrix $A' = \begin{pmatrix} A &amp; X \cr 0 &amp; Y \end{pmatrix}$ is unitary. So consider the operators $B' = \begin{pmatrix} B &amp; 0 \cr 0 &amp; 0 \end{pmatrix}$ and $C' = \begin{pmatrix} C &amp; 0 \cr 0 &amp; 0 \end{pmatrix}$ on $H \oplus K$. The operators $B'$ and $C'$ are partial isometries on $H \oplus K$, so by the work above, $$P(P(A'B')C') = P(A' P(B'C')).$$ Now calculate: $A'B' = \begin{pmatrix} AB &amp; 0 \cr 0 &amp; 0 \end{pmatrix}$ and so a moment's thought ought to show that $P(A'B') = \begin{pmatrix} P(AB) &amp; 0 \cr 0 &amp; 0 \end{pmatrix}$, making the left hand side of the above equal to $$P(\begin{pmatrix} P(AB) C &amp; 0 \cr 0 &amp; 0 \end{pmatrix}) = \begin{pmatrix} P(P(AB) C) &amp; 0 \cr 0 &amp; 0 \end{pmatrix}.$$ The right hand side is almost the same, but with $P(A P(BC))$ in the upper left corner, and unless I made a ridiculous error, you have what you want. (There is a sightly more complicated unitary dilation theorem for operators between different spaces, so if there were no mistakes in this approach, maybe the same idea works in the general case too.)</p> http://mathoverflow.net/questions/45428/moving-to-academia-from-industry/45535#45535 Answer by anon for Moving to academia from industry anon 2010-11-10T08:26:34Z 2010-11-10T08:26:34Z <p>Nobody seems to have mentioned much about teaching--- perhaps because the original question itself makes no mention of teaching having anything to do with the desire to return to academia. This is a kind of elephant in the room. </p> <p>I should admit: I'm on the academic side, I have not personally tried to make this kind of transition, and I have never been in a position to evaluate somebody making this kind of transition. But it seems to me that if you're reasonably current with your research area, and publishing papers, and meeting people (as suggested elsewhere), your biggest obstacle may be teaching.</p> <p>Presumably you have no teaching experience over the last n years, and depending on your grad school experience, you may not have had much then (or it may have been a different sort from what professors do). This may matter. I don't know how to begin building a teaching history, while working a full-time job.</p> <p>You may need to overcome the suspicion that will find teaching low-level service courses boring for the same reasons you find your current job in industry boring. Imagine the skeptic on the search committee who asks, rhetorically, "Who <em>wouldn't</em> be an academic if it were all just learning, writing papers, and talking to enthusiastic people with the same interests?"</p> <p>Even with stellar references and a personal connection or three in the department, someone will ask: can you teach? Do you want to? What's the answer, and how do you convey it on your CV? </p> <p>I don't have specific advice in this area, because it depends on where you want to work, and your own background. If it is possible to do pedagogical things in your current job, or service/outreach to non-specialists or students, perhaps that would help. Maybe actual teaching (on a per-course basis, not as tenure-track faculty) or volunteering would help. My feeling is that you need to do something to address these issues head-on, to confront both any genuine gaps in your CV, and the biases and prejudices you may face simply because you are changing careers.</p> http://mathoverflow.net/questions/44907/an-example-of-a-non-paracompact-tvs-over-the-reals-say/44929#44929 Answer by anon for An example of a non-paracompact tvs (over the reals, say) anon 2010-11-05T07:37:30Z 2010-11-05T08:09:25Z <p>If I have not goofed on some detail, here is an example. Let $w$ be the first uncountable ordinal (= the set of countable ordinals). Regarded as a topological space with its usual order topology, it is not paracompact. The space $\mathbb R^w$ with the usual product topology is a topological vector space over $\mathbb R$ in an obvious way. Let $V$ denote the subspace $\mathbb R^w$ consisting of those functions from $w$ to $\mathbb R$ whose support is at most countable. Then $V$ is a topological vector space over R in an obvious way. $V$ contains a closed subset homeomorphic to $w$ (namely the set of functions $g_x$, where $g_x(y) = 1$ if $y &lt; x$ and $0$ otherwise; the map sending $x$ in $w$ to $g_x$ is a homeomorphism onto its range). A closed subspace of a paracompact space must be paracompact, so $V$ is not paracompact.</p> http://mathoverflow.net/questions/44853/simplest-proof-of-dimension-of-solution-space-for-linear-odes/44916#44916 Answer by anon for Simplest proof of dimension of solution space for linear ODEs anon 2010-11-05T02:35:24Z 2010-11-05T02:35:24Z <p>I don't know of an easy/easiest way to prove it for a general $n$th degree linear ODE, but it is worth pointing out that in the constant coefficient case you can get this from elementary linear algebra. The idea is that if $N$ is a positive integer and you have complex numbers $c_1, \dots, c_N$, then the solutions to the differential equation $$\sum_{n=0}^N c_k y^{(k)} = 0$$ (here $y^{(k)}$ denotes the $k$th derivative of $y$, interpreted as $y$ when $k=0$) are precisely the elements of the kernel of the operator $$T = \sum_{n=0}^N c_k D^k$$ where $D$ is differentation, regarded as an operator on a vector space $V$ of functions (there is some freedom in what particular space you choose here; say the set of all infinitely differentiable functions $\mathbb{R} \to \mathbb{C}$). From the fundamental theorem of algebra, you know there are complex numbers $\omega, \omega_1, \dots, \omega_N$ with the property that the polynomial $\sum_{n=0}^N c_k z^k$ factors as $\omega \prod_{n=1}^n (z - \omega_n)$; it follows that your operator $T$ also factors, in the algebra of operators on $V$, as $$T = \omega \prod_{n=1}^N (D - \omega_n I),$$ where $I$ denotes the identity operator on $V$. </p> <p>The point is that each of the operators $D - \omega_n I$ has a one-dimensional kernel by basic calculus. (For any $k$, the function $f(t) = \exp(kt)$ is a solution to $y' = k y$, and if $g$ is any other, the quotient rule for derivatives shows that $(g/f)' = 0$. So by a standard argument involving the mean value theorem, $g/f$ is constant; so ${f}$ is a basis for $D - kI$.)</p> <p>And it is a basic linear algebra fact that a product of $n$ operators with one-dimensional kernel, can have kernel of dimension at most $n$. (Follows from the more general assertion that if $S_1: V \to V$ and $S_2: V \to V$ are any operators, the dimension of the kernel of $S_1 S_2$ is at most the dimension of the kernel of $S_1$ plus the dimension of the kernel of $S_2$. This very easy consequence of the rank-nullity theorem--- and does not require $V$ to be finite dimensional.)</p> <p>Why is the kernel of $T$ exactly $n$-dimensional? Well, just write down $n$ linearly independent elements in it, as they do in textbooks. (Of course, if you have the better sort of textbook, the entire argument just given is in there.)</p> <p>For non-constant coefficients, factoring the corresponding differential operator is no longer the way you want to approach this. But for a lot of ODE, you can still get reasonably elementary theorems about the dimension of the kernel of the operator by applying some kind of transform (e.g. the Laplace transform) and getting in a position where it is just algebra again. </p> http://mathoverflow.net/questions/43755/when-should-you-and-should-you-not-share-your-mathematical-ideas/43940#43940 Answer by anon for When should you, and should you not, share your mathematical ideas? anon 2010-10-28T05:23:51Z 2010-10-28T05:23:51Z <p>Thus far, the responses about the negative aspects of sharing only discuss idea "theft." This would seem to apply mostly when the ideas being shared are already publishable, near publishable, or at least very likely to lead to something publishable. The biggest danger I have ever experienced with sharing occurs earlier in the process and has nothing to do with theft. It is to enter the following pattern:</p> <p>(1) I share an idea or problem with somebody before I have really given myself time to marinate in it. (2) The other person introduces their own ideas--- and is so persuasive, or detailed, or optimistic about their own views, that I get caught up in them. (3) I lose the thread of my original idea and/or spend a long time fruitlessly trying to attack my own problem with somebody else's way of thinking.</p> <p>I'm not saying that my way is always better or leads to the right thing, and that the only thing stopping me from publishing great results is distractions from lesser minds. I am saying that sometimes you should not share your ideas until you are either well and truly stuck, or far enough along that you could give a coherent (if detail-free) seminar talk about them. You need to know enough about your own ideas that you can put them down for a minute and compare them with someone else's without information loss. When this is, exactly, changes from person to person and problem to problem. But if you share too soon, you run a risk of spending a lot of mathematician-hours checking the details of somebody else's half-baked idea before you have even checked your own. I think grad students are generally more in danger of this than they are of idea theft. (It is easy to unconsciously take an established person's educated guess for "expert wisdom", even when it is not intended as such.)</p> http://mathoverflow.net/questions/42336/name-for-a-basic-principle-of-calculus/42348#42348 Answer by anon for Name for a basic principle of calculus? anon 2010-10-16T00:27:36Z 2010-10-16T00:27:36Z <p>I second the idea that this, or something like this, is a version of Green's theorem. (For a family of simple closed curves c_t, anyway, a precise statement would replace "size of boundary times rate of motion of boundary" with the integral of the normal component of the flow velocity (ie the vector you get by differentiating c_t with respect to t, holding the curve variable fixed) over c_t. And the simplest--- maybe only--- way I can think of to prove that is to use Green's theorem to write the area enclosed by c_t as a contour integral.)</p> http://mathoverflow.net/questions/39538/what-is-the-naming-reason-of-poles-in-complex-analysis/39547#39547 Answer by anon for What is the naming reason of poles in complex analysis? anon 2010-09-21T23:00:00Z 2010-09-21T23:00:00Z <p>This may be apocryphal folk etymology, but I always thought it was because if you plot, or envision plotting, the surface z = |f(x + iy)|, at poles of f, the surface, if you imagine it sitting over the xy plane, looks like it is being supported by a really tall pole. Like a circus tent. I have no citations to support this belief, but I must have gotten it from somewhere. Anyway it makes a good deal of sense. </p> <p>I'm posting this, despite not having an MO account, because genuinely can't understand why nobody has posted it yet. (Nikita's "because poles stick up" comes close, but seems to have been drowned out by posts about poles being "big", or invocations of the north pole, which seem to be entirely different explanations.)</p> http://mathoverflow.net/questions/62629/textbook-recommendations-for-undergraduate-proof-writing-class/62681#62681 Comment by anon anon 2011-04-26T00:17:11Z 2011-04-26T00:17:11Z I recommend anything but this book. When I taught out of it, prepared students liked it a lot, but borderline students (who need such a class more) struggled more than they might have without any book. It <i>hurts</i> weak students. It seems to feed an expectation in them that all proof-writing is done &quot;step-by-step&quot;, with the precise sequence of steps dictated entirely by the formal structure of the statement to be proved, and the exercises do not carefully delineate between what math can be taken for granted and what cannot--- giving the impression that &quot;proof writing&quot; can be done in a vacuum. http://mathoverflow.net/questions/62629/textbook-recommendations-for-undergraduate-proof-writing-class/62878#62878 Comment by anon anon 2011-04-26T00:01:03Z 2011-04-26T00:01:03Z I have taught out of this book and second the recommendation. (But yeah, the price.) http://mathoverflow.net/questions/62866/recent-applications-of-mathematics/62875#62875 Comment by anon anon 2011-04-25T07:49:14Z 2011-04-25T07:49:14Z FWIW I believe linking-to-the-arXiv orthodoxy is to link to the main page of the article and not the PS_cache copy of a particular version of it. <a href="http://arxiv.org/abs/0809.2079" rel="nofollow">arxiv.org/abs/0809.2079</a> http://mathoverflow.net/questions/1083/do-good-math-jokes-exist/1229#1229 Comment by anon anon 2011-04-18T00:08:08Z 2011-04-18T00:08:08Z Just wondering: if you were wearing a skirt in the same room that somebody was saying one of these, how would you feel? The attributions to these jokes suggest that they are decades old, and it shows. In my opinion, these jokes should be left in those decades. http://mathoverflow.net/questions/60255/information-about-publishing-and-citations/60256#60256 Comment by anon anon 2011-04-01T05:04:42Z 2011-04-01T05:04:42Z A caveat with this (or any MSC data) is that most authors self-classify their articles, or editors do it, and there is <i>huge</i> variation in how. Personally I narrow my work down to a few MSCs, but which ones is almost random; I never put <i>all that apply</i>. No two MSC topics are conceptually disjoint (e.g. operator/systems theory, Fourier/abstract harmonic analysis, CS/logic, category theory/everything), so the precise methodology in going from &quot;MSC hits for X&quot; to &quot;papers/activity in X&quot; seems more significant than the data itself. That Grcar did not address this is a huge hole in his argument. http://mathoverflow.net/questions/50085/a-comparison-on-matrix-transpose-question Comment by anon anon 2010-12-21T22:42:24Z 2010-12-21T22:42:24Z If you think the question is illegitmate, is it too much to ask that you vote to close or downvote and stop there? Why use comments as an passive-aggressive outlet for your suspicions and what you find hard to believe? You don't <i>have to</i> comment on a question. And as for &quot;unwilling to give his or her name&quot;, why would anyone new to MO register under their real name if their questions are treated like this? http://mathoverflow.net/questions/40729/does-a-referee-have-to-check-carefully-the-proof Comment by anon anon 2010-12-12T23:54:56Z 2010-12-12T23:54:56Z This is only somewhat related, so it is a comment. There is clearly some variation of opinion on the extent to which a referee's job is to check proofs, or to which a referee/editor bears responsibility for errors in published work, and so on. But whatever your opinion: if you decide to reject a paper on the grounds that its results are largely contained in previously published work, <i>you should make reasonably sure that the work you refer to is free from error</i>. At the very least, check Mathscinet to see if the journal has posted a retraction. Sadly, I speak from multiple experiences. http://mathoverflow.net/questions/48818/on-common-fixed-points-of-commuting-polynomials-and-rational-functions Comment by anon anon 2010-12-09T23:33:28Z 2010-12-09T23:33:28Z You might like Shields' 1964 Proc AMS paper &quot;On fixed points of commuting analytic functions&quot; where he shows that if f and g are commuting, analytic on the unit disc, and nice on the boundary, then f and g have a common fixed point. A 1973 Proc AMS paper of Behan &quot;Commuting analytic functions without fixed points&quot; and a 1984 Trans AMS paper of Cowen &quot;Commuting analytic functions&quot; have more results in this line. (Some of the conjectures in this last paper are disproved in Chalendar and Mortini's more recent &quot;When do finite Blaschke products commute?&quot;). Rational functions are subtle things. http://mathoverflow.net/questions/47850/associativity-of-polar-decomposition/48681#48681 Comment by anon anon 2010-12-09T07:57:34Z 2010-12-09T07:57:34Z Wow, neat. I wonder if this is a &quot;minimal&quot; example (or if it's not, what a minimal example looks like). My own search was with smaller matrices than this (everything between spaces of dimension at most 3), and nothing seemed to work, so I gave up. http://mathoverflow.net/questions/29323/math-puzzles-for-dinner/34000#34000 Comment by anon anon 2010-12-08T00:26:20Z 2010-12-08T00:26:20Z I wonder if this problem was inspired by Edgar Allan Poe's &quot;The Gold-Bug,&quot; which contains a search for a treasure on an island using a tree and a specified direction. The search does not require any math (although the message telling them where to search is written in code, and the story explains how to solve a simple substitution cipher). I imagine a mathematically inclined reader liking the story, but wanting to improve how Poe's character hid the treasure. I bet Poe would have liked this problem. http://mathoverflow.net/questions/47850/associativity-of-polar-decomposition/47887#47887 Comment by anon anon 2010-12-02T00:18:07Z 2010-12-02T00:18:07Z Ack! I even got the term wrong (my &quot;dilation&quot; is really an extension, and as you point out, things with kernels can't extend to things that don't have them). One <i>can</i> dilate (in the proper sense of the term) any contraction to a unitary, but the $A'$'' you get will have to have some junk in the lower left corner. Perhaps it is possible to salvage this idea by understanding what happens with the junk when you do a polar decomposition. But maybe it is no simpler than trying another approach. (I leave my whole answer unedited so that these comments make sense to people who read them.) http://mathoverflow.net/questions/45428/moving-to-academia-from-industry/45525#45525 Comment by anon anon 2010-11-10T07:24:07Z 2010-11-10T07:24:07Z I'd add: do joint work (although it practically follows, logically, from &quot;meet people,&quot; it is worth emphasizing). Successful and ongoing collaborations with active mathematicians would go a longer way at most places, I think, than the same number of solo papers. Or even twice as many. Although it occurs to me: for many of the same reasons that it is unwise to talk publically about leaving one's career, it may be unwise to begin doing too much mathematician stuff without an actual job offer. Presumably the OP knows what risks there are (if any) in their specific situation. Good luck! http://mathoverflow.net/questions/44853/simplest-proof-of-dimension-of-solution-space-for-linear-odes Comment by anon anon 2010-11-05T05:19:55Z 2010-11-05T05:19:55Z I don't understand why this was closed. I'd bet if someone asked for the easiest proof of some standard result in another subject (e.g. Bezout's theorem for plane algebraic curves), there would be a host of responses. Mathematicians do interest themselves in questions like these, especially when teaching. I can't help but suspect that because the subject matter is differential equations, and not something nearer and dearer to the community's heart, the benefit of the doubt is not being given, and it is being assumed that this is some kind of homework thing and not a valid question. http://mathoverflow.net/questions/44853/simplest-proof-of-dimension-of-solution-space-for-linear-odes/44916#44916 Comment by anon anon 2010-11-05T05:10:41Z 2010-11-05T05:10:41Z If the OP isn't reading that closely, what would a helpful answer look like? http://mathoverflow.net/questions/43341/technical-phd-after-33 Comment by anon anon 2010-10-24T00:00:33Z 2010-10-24T00:00:33Z I don't think it a good idea to get a PhD &quot;in the hope of becoming a professor&quot; at any age. The jobs you get post-PhD depend on factors that are completely unknown going in: your research, your teaching, your connections, your own feelings about jobs, the state of the market when you finish. I wouldn't invest the time in the hope of a specific job outcome. I'd recommend it only if you'd still want it in a world where careers didn't exist. This may be overly romantic, but the alternative doesn't seem very practical, either. Since I'm answering a different question, this is a comment.