User orr shalit - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T03:34:33Z http://mathoverflow.net/feeds/user/1193 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4699/examples-of-left-reversible-semigroups Examples of left reversible semigroups Orr Shalit 2009-11-09T01:18:28Z 2011-10-27T02:24:56Z <p>I am looking for concrete examples of cancellative, left reversible semigroups. Left reversible semigroups are also called "Ore semigroups". See <a href="http://en.wikipedia.org/wiki/Special%5Fclasses%5Fof%5Fsemigroups" rel="nofollow">this wikipedia page</a> for the definition of a left reversible semigroup. Of course, commutative semigroups are automatically left reversible, and I am looking for non-commutative examples.</p> <p>Please also mention if these semigroups arise in an interesting setting. </p> http://mathoverflow.net/questions/70866/question-on-linear-operators/70872#70872 Answer by Orr Shalit for Question on Linear Operators Orr Shalit 2011-07-21T03:00:42Z 2011-07-21T03:00:42Z <p>$exp(L)$ is bounded, regardless of the local nilpotentcy, since $\|L^n\|\leq \|L\|^n$. On the other hand, if you wanted to ask the question about unbounded $L$ (say, for all $v$ in the domain), then the answer is no.</p> http://mathoverflow.net/questions/1684/why-is-the-exterior-algebra-so-ubiquitous/68153#68153 Answer by Orr Shalit for Why is the exterior algebra so ubiquitous? Orr Shalit 2011-06-18T15:24:09Z 2011-06-18T15:24:09Z <p>I think that what unifies some of the different examples of when the exterior algebra occurs is that it is the structure that transforms the action of a commutative ring on a module (or, more concretely, the action of several commuting linear operators on a vector space) into a chain complex. The ubiquitous structure is really <em>commutative</em> rings and modules over commutative rings. The exterior algebra actually encodes commutativity, in a sense. (One might then ask why is it that mathematicians love so much commutative rings, commuting operators, etc.)</p> <p>When tensoring a commutative action with the graded algebra one gets the Koszul complex, and the anti-commutative nature of the graded algebra is precisely what, when coupled with the commutative action of the ring, makes possible the definition of a dirac operator $d$ with $d^2=0$.</p> <p>I got these ideas after thinking about Joseph Taylor's paper, "A joint spectrum for several commuting operators", J. Funct. Anal. , 1970.</p> http://mathoverflow.net/questions/65394/there-must-be-a-good-introductory-numerical-analysis-course-out-there/67720#67720 Answer by Orr Shalit for There must be a good introductory numerical analysis course out there! Orr Shalit 2011-06-14T02:45:57Z 2011-06-14T02:45:57Z <p>If you haven't, go to the library and take a look at this book: "Numerical Analysis: A Mathematical Introduction", Michelle Schatzman.</p> <p>It will give you some ideas how to make students fall in love with numerical analysis.</p> http://mathoverflow.net/questions/44705/cardinalities-larger-than-the-continuum-in-areas-besides-set-theory/44766#44766 Answer by Orr Shalit for Cardinalities larger than the continuum in areas besides set theory Orr Shalit 2010-11-04T02:07:13Z 2010-11-04T02:07:13Z <p>Denote by $H^\infty$ the Banach algebra of all bounded analytic functions on the unit disc. Then the <strong>maximal ideal space</strong> of $H^\infty$, (i.e., the set of all norm continuous, multiplicative linear functionals on $H^\infty$) has cardinality $2^{2^{\aleph_0}}$. SInce maximal ideal spaces are used all the time in Banach algebra theory, this is an example of a naturally occuring large cardinal". </p> <p>This is also an example of how knowing the (large) cardinality of a set may help direct us in research, even if this knowledge does not lead to a proof of anything. In this example, since the maximal ideal space is so wild, one realizes that it is hopeless" to find a nice analytic or geometric description of the maximal ideal space, and one is led to look for a suitable replacement. In algebras such as $H^\infty$ it is often the weak-$*$ continuous multiplicative linear functionals which are more useful (there are only $2^{\aleph_0}$ such functionals).</p> http://mathoverflow.net/questions/36596/refereeing-a-paper/41677#41677 Answer by Orr Shalit for Refereeing a Paper Orr Shalit 2010-10-10T13:30:01Z 2010-10-10T14:12:16Z <p>The referee <strong>must</strong> check if the result is correct! That is, the referee should check the <strong>proofs</strong>. This point seems to be so under-estimated, but in my opinion it is the most important part of the referee's job. The referee is the only one who will do this, except the author. There does need to be a second check of correctness after the author. There are many flawed papers out there. This has cost many other researchers time and health.</p> <p>If the referee is unsure, he should ask the author for help in understanding the proofs (and the author should be helpful). It can be a change in the paper or a note for the referee.</p> <p>As for whether or not the paper is "interesting": I decline to referee a paper if it is not interesting (in a broad sense) enough for me to put the effort in checking whether or not it is correct. Thus if all referees were correctness minded, then "interesting" papers would get their advantage.</p> http://mathoverflow.net/questions/41041/if-two-homogeneous-algebraic-varieties-are-isomorphic-are-they-necessarily-rel If two "homogeneous" algebraic varieties are isomorphic, are they necessarily related by a linear map? Orr Shalit 2010-10-04T17:50:01Z 2010-10-10T01:39:10Z <p>Let $I,J$ be homogeneous ideals in the algebra of polynomials in $n$ variables over the complex numbers. Let $V(I)$ and $V(J)$ be the <strong>affine</strong> algebraic varieties that are determined by $I$ and $J$ (<strong>not</strong> the projective varieties). Suppose that $V(I)$ and $V(J)$ are isomorphic as algebraic varieties. By this I mean that there are polynomial maps $f$ and $g$ from $\mathbb{C}^n$ to itself, such that $f$ restricted to $V(I)$ is a bijection onto $V(J)$, and such that $g$ restricted to $V(J)$ is its inverse.</p> <p>The question is this: does it follow that there exists a linear map on $\mathbb{C}^n$ that maps $V(I)$ onto $V(J)$?</p> <p>Thanks to discussions with colleagues (thank you David and Mike), I am quite convinced that if we assume that the origin is the only singular point in $V(I)$ then the answer is yes. Is this true in general?</p> <p>I think this question is equivalent to the following (see my partial answer below): <strong>Is it true that whenever there is an isomorphism between $V(I)$ and $V(J)$, there is also isomorphism that fixes $0$?</strong></p> http://mathoverflow.net/questions/41041/if-two-homogeneous-algebraic-varieties-are-isomorphic-are-they-necessarily-rel/41094#41094 Answer by Orr Shalit for If two "homogeneous" algebraic varieties are isomorphic, are they necessarily related by a linear map? Orr Shalit 2010-10-05T01:35:31Z 2010-10-07T23:48:58Z <p>I'll share here two proofs of the fact I am interested in under the assumption that the origin is the only singular point. </p> <p>Proof 1: (This is equivalent I think to Torsten's answer, and was a explained to me by a colleagues - thank you David and Mike) Let $F: V(I) \rightarrow V(J)$ be an isomorphism. Then since $0$ is the only singular point in $V(I)$, then it is mapped to $0 \in V(J)$. Now the derivative of $F$ at $0$, call it $DF$, is a linear map in $\mathbb{C}^n$ that takes the tangent cone of $V(I)$ at $0 \in V(I)$ to the tangent cone of $V(J)$ at $0 \in V(J)$. But these tangent cones are $V(I)$ and $V(J)$, respectively. Thus, $DF$ is the required linear map taking $V(I)$ onto $V(J)$.</p> <p>Proof 2: There are some technical details missing here. Let $F: V(I) \rightarrow V(J)$ be an isomorphism. Again, $F$ must take $0$ to $0$, under the assumption that $0$ is the only singular point. Thus, $F$ has the form $$F(z) = Az + \textrm{ higher order terms .}$$ Now define $F_t$ by $$F_t(z) = tF(z/t) .$$ Since $I$ and $J$ are homogeneous, $V(I)$ and $V(J)$ are invariant under scalings, so $F_t$ is again an isomorphismof $V(I)$ and $V(J)$. But $F_t$ has the form $$F_t(z) = Az + \frac{1}{t}(\textrm{higher order terms}).$$ Taking $t \rightarrow \infty$, we converge to the isomorphism (hopefully) $z \mapsto Az$.</p> <p>That's for the case when $0$ is the only singular point. In fact, all that is used is that there is an isomorphism taking $0$ to $0$. Is it true that <strong>whenever there is an isomorphism between $V(I)$ and $V(J)$, there is also isomorphism that fixes $0$?</strong> (here $I$ and $J$ are assumed homogeneous, of course).</p> http://mathoverflow.net/questions/3249/is-there-a-stable-algorithm-for-polynomial-division-in-several-variables/5223#5223 Answer by Orr Shalit for Is there a stable algorithm for polynomial division (in several variables)? Orr Shalit 2009-11-12T18:46:40Z 2010-06-25T04:27:20Z <p><strong>Revised June 25, 2010: I feel I need to close this circle. Thanks for the answers given but they have all been quite off the mark. My partial results on this problem (there is much room for improvement) as well as an application to operator theory can be found in this link:</strong> <a href="http://arxiv.org/abs/1003.0502" rel="nofollow">http://arxiv.org/abs/1003.0502</a></p> http://mathoverflow.net/questions/3249/is-there-a-stable-algorithm-for-polynomial-division-in-several-variables Is there a stable algorithm for polynomial division (in several variables)? Orr Shalit 2009-10-29T12:09:30Z 2010-06-25T04:27:20Z <p>Suppose you have a homogeneous ideal $I$ inside the algebra $\mathbb{C}[x_1,...,x_d]$ of complex polynomials in $d$-variables. Can one find a basis for $I$, say ${f_1,...,f_k}$, such that every $h \in I$ can be written as </p> <p>$$h = a_1 f_1 + ... + a_k f_k$$</p> <p>where the coefficients appearing in each summand $a_i f_i$ are not much bigger then the coefficients appearing in $h$? More specifically, given that ${f_1,...,f_k}$ is a Groebner basis for $I$, can one modify the standard division algorithm so that one gets $h = a_1 f_1 + ... + a_k f_k$ with controlled terms?</p> <p><strong>Added 13.11.09 -</strong> By <em>controlled</em> I mean that the <strong>coefficients</strong> of the terms $a_i f_i$ are bounded in a non-exponential manner by the coefficients of $h$. There is no problem with <strong>degree</strong> of the $a_i$'s.</p> <p>I will share that I found this possible in some special cases, for example when $d=2$ or when $I$ is generated by monomials, and I am now interested in the general question.</p> <p>Note: My question begins <strong>after</strong> a basis has been found, I am <em>not</em> concerned here with the terrible complexity of actually computing a Groebner basis.</p> <p><strong>Another note (added 12.11.09): The answers and links that I am getting suggest that this problem has not been considered before. So I re-eamphasize my note from above: assume that a Groebner basis, even a universal Groebner basis, has already been found for the ideal. What can be said about the stability of certain variants of the division algorithm now?</strong> </p> http://mathoverflow.net/questions/2556/real-world-applications-of-mathematics-by-arxiv-subject-area/2756#2756 Answer by Orr Shalit for Real-world applications of mathematics, by arxiv subject area? Orr Shalit 2009-10-27T03:33:36Z 2010-01-28T20:03:06Z <p><strong>math.OA Operator Algebras</strong></p> <ul> <li>(In "Nature") Operator algebras, and, more broadly, operator theory, appear in mathematical models for quantum phenomena. </li> <li>(In Engineering) Completely positive maps are <a href="http://en.wikipedia.org/wiki/Quantum%5Fchannel" rel="nofollow">used</a> in quantum information theory. There are also many <a href="http://books.google.com/books?id=CKVXzNo6KwwC" rel="nofollow">connections between operator algebras and wavelets</a>, which is useful in electrical engineering.</li> </ul> http://mathoverflow.net/questions/4994/fundamental-examples/6462#6462 Answer by Orr Shalit for Fundamental Examples Orr Shalit 2009-11-22T14:02:29Z 2009-11-22T21:48:51Z <p>In <code>$C^*$</code>-algebras, the <a href="http://en.wikipedia.org/wiki/Cuntz%5Falgebra" rel="nofollow">Cuntz Algebra</a> is a fundamental example of a separable unital <code>$C^*$</code>-algebra. Its appearance has reshaped much of the theory of <code>$C^*$</code>-algebras.</p> http://mathoverflow.net/questions/6423/set-theory-for-category-theory-beginners/6460#6460 Answer by Orr Shalit for Set theory for category theory beginners Orr Shalit 2009-11-22T13:12:58Z 2009-11-22T21:38:06Z <p>I recommend "Set Theory", by Pinter. It is a very concise book, and if I am not mistaken, it uses basically von Neumann's approach to classes and sets (you asked for NBG class theory). You will only need to read the relevant (short) chapter there to feel very comfortable with sets and classes. And if you will want to learn about cardinality issues - you'll find it there in concise form.</p> <p>This is not the book I would reccommend to someone who knows no math and would like to learn set theory, but for you I recommend it.</p> http://mathoverflow.net/questions/4994/fundamental-examples/5346#5346 Answer by Orr Shalit for Fundamental Examples Orr Shalit 2009-11-13T13:53:29Z 2009-11-13T13:53:29Z <p>In operator theory, the <a href="http://en.wikipedia.org/wiki/Shift%5Foperator" rel="nofollow">unilateral shift</a>. This operator is not only the fundamental example of an isometry on a Hilbert space, it can also be shown that this operator contains all other operators, in some sense. </p> http://mathoverflow.net/questions/5074/are-there-elementary-school-curricula-that-capture-the-joy-of-mathematics/5082#5082 Answer by Orr Shalit for Are there elementary-school curricula that capture the joy of mathematics? Orr Shalit 2009-11-11T18:07:04Z 2009-11-11T18:07:04Z <p>Hello! I am mathematician and a homeschooler (that is - my children are homeschooled). If you want to capture the joy of mathematics, or the joy of anything for that matter, then it is my personal belief that the best curriculum is no curriculum. That last sentence is a bit of an exaggeration, but still my experience has shown that the most important things children discover themselves, and the most joyful moments of discovery are when one discovers something for oneself. Since you emphasize joy in your question, this is my answer.</p> http://mathoverflow.net/questions/4699/examples-of-left-reversible-semigroups/4920#4920 Answer by Orr Shalit for Examples of left reversible semigroups Orr Shalit 2009-11-10T20:50:44Z 2009-11-10T20:50:44Z <p>Here is an example that I found in the book "Algebraic Theory of Semigroups, vol. I" by Clifford and Preston (the exercise on p. 36): The universal semigroup generated by two elements $a,b$ such that $ab = ba^k$. This semigroup can be concretely described as the set of of pairs $(i,j)$, with $i,j$ nonnegative integers, with multitplication $$(i,j)(m,n) = (i+m, k^m + n) .$$</p> http://mathoverflow.net/questions/4137/one-parameter-semigroups-of-bimodules/4211#4211 Answer by Orr Shalit for One-parameter semigroups of bimodules Orr Shalit 2009-11-05T04:09:17Z 2009-11-05T05:08:53Z <p>Such structures have been investigated at depth (welcome to the club!). Let me try to answer some of your questions.</p> <p>1) First, let me suggest that you look at Bill Arveson's book on this subject, <a href="http://www.springer.com/math/analysis/book/978-0-387-00151-7" rel="nofollow">Noncommutative Dynamics and E-semigroups</a>. Arveson is considered the pioneer of modern work on these structures, and you will enjoy his book.</p> <p>2) An "infinitesemal generator" can mean various things, but in some sense, remarkably, it doesn't always exist. Arveson's book contains some treatment of this issue, known as the existence of type III examples, which are due to R. T. Powers and B. Tsirelson (there is also recet work of M. Izumi).</p> <p>3) Do we need continuity conditions? Measurability conditions suffice. The condition is that the bundle {E_t}_{t>0} be isomorphic, as a measurable bundle of Hilbert spaces, to the trivial bundle (0,infinity) X H_0, with H_0 some fixed Hilbert space, plus compatibility of the measurable structure with addition, multiplication, etc.</p> <p>4) Extensive research has been carried out also in the case where the E_t are Hilbert bimodules (C^*-correspondences). Search the ArXiv for works of Michael Skeide, or Paul Muhly and Baruch Solel.</p> <p>5) I should also mention: these product systems arise naturally in the study of, give rise to, and are in a one-to-one correspondence with semigroups of *-endomorphisms on von Neumann algebras.</p> http://mathoverflow.net/questions/1282/is-every-norm-in-rn-a-continuous-function/2761#2761 Answer by Orr Shalit for Is every norm in R^n a continuous function? Orr Shalit 2009-10-27T04:05:39Z 2009-10-27T04:05:39Z <p>The answer is "yes". </p> <p>The exaplanation depends on what is meant by "continuous". Let's agree that "the norm || || is continuous" means "if (x_n) is a sequence of vectors such that the coordinates x_n converge to the coordinates of some vector x, then ||x_n|| converges to ||x||".</p> <p>Let e_1, ..., e_n be the standard basis of R^n. First one shows that for every i, the function t -> ||t*e_i|| is continuous (t real). Then, after writing every vector as a_1 * e_1 + ... + a_n * e_n, the result follows easily from the triangle inequality. </p> http://mathoverflow.net/questions/65394/there-must-be-a-good-introductory-numerical-analysis-course-out-there/67720#67720 Comment by Orr Shalit Orr Shalit 2011-06-18T11:32:11Z 2011-06-18T11:32:11Z Sorry, I do not. I just thought it would help to have this at your side if you are going to construct such a course. http://mathoverflow.net/questions/36596/refereeing-a-paper/41677#41677 Comment by Orr Shalit Orr Shalit 2010-10-11T14:07:43Z 2010-10-11T14:07:43Z Perhaps &quot;judgemental&quot; is bad wording. When you say a proof is not correct there is much less room for debate, usually. But saying that a result is not important is something that cannot be proved in the usual way we prove things. http://mathoverflow.net/questions/36596/refereeing-a-paper/41677#41677 Comment by Orr Shalit Orr Shalit 2010-10-10T14:14:39Z 2010-10-10T14:14:39Z You can ask for clarifications via the editor (in some journals at least). Anyway, when refereeing is less judgemental (&quot;this paper is correct but the results are not important&quot;), as I see it, anonimity is less important. http://mathoverflow.net/questions/41041/if-two-homogeneous-algebraic-varieties-are-isomorphic-are-they-necessarily-rel/41509#41509 Comment by Orr Shalit Orr Shalit 2010-10-09T12:51:56Z 2010-10-09T12:51:56Z Thanks! Let me see if I get this right. You are saying that an affine variety $V = V(I)$ (I homogeneous) is always of the form $\mathbb{A}^d \times C(W)$, where $W$ is a projective variety, and that this $\mathbb{A}^d$ is a geometric invariant. Thus, translations along this space leave $V$ invariant, so we get back the case $0$ goes to $0$. I still do not understand why every $V$ has this form. Regarding this, another question: one can look at the singular locus of $V$, $S(V)$, and then look at $S(S(V))$ and so on, until you get a subspace. I this subspace the $\mathbb{A}^d$ from above? http://mathoverflow.net/questions/41041/if-two-homogeneous-algebraic-varieties-are-isomorphic-are-they-necessarily-rel/41047#41047 Comment by Orr Shalit Orr Shalit 2010-10-04T22:55:20Z 2010-10-04T22:55:20Z Thank you for the answer. I understand the first part. It would be interesting to know if the requirement that the cone point be the only the <b>only</b> singular point be removed completely. I do not understand the addendum - is there perhaps an elementary explanation? http://mathoverflow.net/questions/41041/if-two-homogeneous-algebraic-varieties-are-isomorphic-are-they-necessarily-rel Comment by Orr Shalit Orr Shalit 2010-10-04T18:03:39Z 2010-10-04T18:03:39Z I apologize, the question was saved before I completed typing. It is now a complete question. http://mathoverflow.net/questions/6423/set-theory-for-category-theory-beginners/6445#6445 Comment by Orr Shalit Orr Shalit 2009-11-22T14:13:02Z 2009-11-22T14:13:02Z I certainly disagree with the above answer, but I do think that there is room for different opinions in these matters. http://mathoverflow.net/questions/3249/is-there-a-stable-algorithm-for-polynomial-division-in-several-variables/5170#5170 Comment by Orr Shalit Orr Shalit 2009-11-13T11:36:08Z 2009-11-13T11:36:08Z No, it isn't. There is no problem with the degree of the $a_i$'s, when the $f_i$'s are a Groebner basis (with respect to a grlex for instance) the degree of the $a_i$'s will be less than the degree of $h$ (the point of 3.1 is that the generating polynomials there are not assumed to be a basis). The bad thing that can happen, see the link below (Example 2.5), is that h can be a polynomial with very small coefficients, while when the division algorithm is run naively, one gets the the <b>coefficients</b> of the polynomials $a_i f_i$ are huge. http://mathoverflow.net/questions/3249/is-there-a-stable-algorithm-for-polynomial-division-in-several-variables/5170#5170 Comment by Orr Shalit Orr Shalit 2009-11-12T23:30:39Z 2009-11-12T23:30:39Z Thanks, that is nice reference, but if I am not mistaken, it does not address the problem. http://mathoverflow.net/questions/3249/is-there-a-stable-algorithm-for-polynomial-division-in-several-variables Comment by Orr Shalit Orr Shalit 2009-11-12T02:29:53Z 2009-11-12T02:29:53Z The answer below was interesting but did not answer my question. I set a bounty in hope it will revive interest. http://mathoverflow.net/questions/4699/examples-of-left-reversible-semigroups Comment by Orr Shalit Orr Shalit 2009-11-12T02:16:33Z 2009-11-12T02:16:33Z Thanks for the answers. I am quite satisfied now. http://mathoverflow.net/questions/4699/examples-of-left-reversible-semigroups/4850#4850 Comment by Orr Shalit Orr Shalit 2009-11-12T02:15:10Z 2009-11-12T02:15:10Z Thanks for the answer. Naturally, I did not know of many concrete examples of Ore left domains, either. http://mathoverflow.net/questions/4699/examples-of-left-reversible-semigroups/4920#4920 Comment by Orr Shalit Orr Shalit 2009-11-11T02:03:54Z 2009-11-11T02:03:54Z Thanks, that's interesting. http://mathoverflow.net/questions/4699/examples-of-left-reversible-semigroups/4794#4794 Comment by Orr Shalit Orr Shalit 2009-11-10T20:39:09Z 2009-11-10T20:39:09Z Thanks. Please don't erase, that's an interesting example and explanation, and somebody might find it useful. http://mathoverflow.net/questions/4187/normal-operators-and-its-spectrum-in-c-algebras/4198#4198 Comment by Orr Shalit Orr Shalit 2009-11-05T03:48:14Z 2009-11-05T03:48:14Z The absolute value of that is a projection, so the spectrum is {0,1}, not [0,1].