User jasomill - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T18:00:27Z http://mathoverflow.net/feeds/user/9352 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/26416/what-is-your-favorite-proof-of-tychonoffs-theorem/65510#65510 Answer by jasomill for What is your favorite proof of Tychonoff's Theorem? jasomill 2011-05-20T08:39:03Z 2011-05-20T08:39:03Z <p>The non-standard analysis proof is an interesting "application" of the ultrafilter proof: a topological space $A$ is compact if and only if every point in the associated "non-standard topological space" ${}^*A$ is <em>near-standard</em>, that is to say, if and only if each $x \in {}^*A$ is contained in every open neighborhood of some standard point $y \in A$ ($i.e.,$ for all $U \subset A$, $U$ open and $y \in U$ implies $x \in {}^*U \subset {}^*A$).</p> <p>So let $\mathcal{X}$ be a set of topological spaces indexed by $I$, and $P$, the product of these spaces; write ${}^*P$ for the "non-standard product" of the set ${}^*\mathcal{X}$ of topological spaces indexed by ${}^*I$, and let $x \in {}^*P$. It suffices to show that $x$ is near-standard.</p> <p>For each $\kappa \in I$, let $x_\kappa \in {}^*X_\kappa \in {}^*\mathcal{X}$ be the $\kappa$th factor of $x$. Then $x_\kappa$ is necessarily near-standard, because $X_\kappa \in \mathcal{X}$ is compact. But this means we can find a point $y \in P$ with factors $y_\kappa \in X_\kappa$ such that $U \subset X_\kappa$ open and $y_\kappa \in U$ implies $x_\kappa \in {}^*U \subset {}^*X_\kappa$, thus $V \subset P$ open and $y \in V$ implies $x \in {}^*V \subset {}^*P$. But this means $x$ is near-standard, so $P$ is compact.</p> <p>"Under the hood," this is basically the ultrafilter proof (my favorite, to answer the original question), so the axiom of choice is required in more or less the same places: while the non-standard objects exist by the Boolean prime ideal theorem, "finding" the $y_\kappa$ in non-Hausdorff spaces requires the full axiom of choice.</p> http://mathoverflow.net/questions/123/linearity-of-the-inner-product-using-the-parallelogram-law/61908#61908 Answer by jasomill for Linearity of the inner product using the parallelogram law jasomill 2011-04-16T10:59:04Z 2011-04-16T10:59:04Z <p>The easiest way to see how things "break" in the case of a more general norm is to look at the shape of its unit "sphere" — unless it's an ellipsoid, no linear transformation exists taking it to a Euclidean sphere, and it follows from the principal axis theorem that each ellipsoid is associated with a unique inner product, and conversely.</p> <p>Quoting Spivak's <em>Comprehensive Introduction to Differential Geometry, Vol. 2</em>, p. 210 (he defines a Minkowski metric to be a map $F: v \to \mathbb{R}$ such that $F(v) > 0$ for all $v \neq 0$ and $F(\lambda v) = | \lambda | F(v)$, so this holds <em>a fortiori</em> under the stronger hypothesis of a norm):</p> <blockquote> <p>THEOREM. Let $F: V \to \mathbb{R}$ be a continuous Minkowski metric on an $n$-dimensional vector space $V$. Suppose that, for all $p$ and $q$ in the unit sphere $\{ v \in V: F(x) = 1 \}$, there is a linear transformation $\phi: V \to V$ such that $\phi(p) = q$ and $F(\phi(v)) = F(v)$ for all $v \in V$. Then $F$ is the norm determined by some positive definite inner product.</p> <p>PROOF: Let $B = \{ v : F(v) \leq 1 \}$, and let $E$ be the unique ellipsoid containing $B$ of smallest volume. Clearly, there must be some point $p$ with $F(p) = 1$ and $p \in$ boundary $E$. Let $q$ be any other point with $F(q) = 1$, and $\phi: V \to V$ a linear transformation with $\phi(p) = q$ such that $F(\phi(v)) = F(v)$ for all $v \in V$. It follows easily from the latter property that $\phi(E) \supset B$. Moreover, $\phi(B) = B$, so $\phi$ is volume preserving. By uniqueness of the ellipsoid $E$, it follows that $\phi(E) = E$. Consequently, $q = \phi(p) \in$ boundary $E$. In other words, every point $q$ with $F(q) = 1$ is in boundary $E$. This means that $E = B$.</p> </blockquote> <p>The required proofs of existence and uniqueness of minimal ellipsoids are, of course, quite easy to motivate geometrically. It's also easy to see why this settles the matter: if isometries don't act transitively on the unit sphere, it's hard to define the "angle between two vectors" in a sensible way.</p> <p>With all that said, a simpler way of looking at things, for me, at least, is this: given any inner product on an $n$-dimensional real vector space, an orthonormal basis exists, in terms of which things are "computationally indistinguishable" from $\mathbb{R}^n$ with the usual inner product — the coefficients don't care what the basis vectors look like. This reduces things to the case of the usual inner product, where "geometric intuition" has been <em>axiomatized:</em> however well-motivated it may be, algebraically, the law of cosines is essentially true <em>by definition</em>.</p> <p>In other words, we always have the option of writing our vectors in a way that makes "ordinary" intuition apply; we even have the option of <em>thinking about things</em> in the usual terms, even when working in a "skew basis." So why complicate matters?</p> <p>Finally, continuity is hardly "un-geometric" in this context: by the triangle inequality, the difference between the lengths of two sides of a triangle is never greater than the length of the third side: $$\Big| \lVert x \rVert - \lVert y \rVert \Big| \leq \lVert x - y \rVert,$$ so any norm on a real vector space is continuous, even Lipschitz.</p> http://mathoverflow.net/questions/54645/what-is-the-best-way-explain-to-undergraduates-that-all-1-dimensional-manifolds-a/54701#54701 Answer by jasomill for What is the best way explain to undergraduates that all 1-dimensional manifolds are orientable? jasomill 2011-02-07T21:57:09Z 2011-02-07T21:57:09Z <p>Elementary? Let M be a curve. Can we find a smooth, non-vanishing vector field defined everywhere on M? This proves the theorem.</p> <p>How to explain? Calling whatever direction the vector points "ahead" is a choice of orientation for the curve.</p> http://mathoverflow.net/questions/5499/which-mathematicians-have-influenced-you-the-most/52280#52280 Answer by jasomill for Which mathematicians have influenced you the most? jasomill 2011-01-17T01:19:34Z 2011-01-17T01:19:34Z <p><strong>Richard Courant</strong>. Several years before I started studying mathematics in earnest, I spent a summer working through his calculus texts. Only recently, on re-reading them, have I come to realize how much my understanding of calculus, linear algebra, and, more generally, of the unity of all mathematics and, to use Hilbert's words, the importance of "finding that special case which contains all the germs of generality," have been directly inspired by Courant's writings.</p> <p>From the preface to the first German edition of his <em>Differential and Integral Calculus:</em></p> <blockquote> <p>My aim is to exhibit the close connexion between analysis and its applications and, without loss of rigour and precision, to give due credit to intuition as the source of mathematical truth. The presentation of analysis as a closed system of truths without reference to their origin and purpose has, it is true, an aesthetic charm and satisfies a deep philosophical need. But the attitude of those who consider analysis solely as an abstractly logical, introverted science is not only highly unsuitable for beginners but endangers the future of the subject; for to pursue mathematical analysis while at the same time turning one's back on its applications and on intuition is to condemn it to hopeless atrophy. To me it seems extremely important that the student should be warned from the very beginning against a smug and presumptuous purism; this is not the least of my purposes in writing this book.</p> </blockquote> <p>Another example: while not a "linear algebra book" <em>per se</em>, I have yet to find a better introduction to "abstract linear algbera" than the first volume of Courant's <em>Methods of Mathematical Physics</em> ("Courant-Hilbert"; so named because much of the material was drawn from Hilbert's lectures and writings on the subject). His one-line explanation of "abstract finite-dimensional vector spaces" is classic: "for <em>n</em> > 3, geometrical visualization is no longer possible but geometrical terminology remains suitable."</p> <p>Lest one be misled into thinking Courant saw "abstract" vector spaces as "$\mathbb{R}^n$ in a cheap tuxedo," he introduces function spaces in the second chapter ("series expansions of arbitrary functions"), and most of the book is about quadratic eigenvalue problems, or, as Courant saw it, "the problem of transforming a quadratic form in infinitely many variables to principal axes."</p> <p>As a final example: Courant's expository <em>What is Mathematics?</em> is perhaps best described as an unparalleled collection of articles carefully crafted to serve as an object at which one can point and say "this is." Moreover, while written as a "popularization," its introduction to constrained extrema problems is, without question, a far, far better introduction than any textbook I've ever seen.</p> <p>I should also mention <strong>Felix Klein</strong>, not only because Klein's views on "calculus reform" so clearly influenced both the style and substance of Courant's texts, but since a number of Klein's lectures have had an equally significant influence on my own perspective. For those unfamiliar with the breadth of Klein's interests, I'm tempted to say "his Erlangen lecture, least of all" (not that there's anything wrong with it).</p> <p>Lest my comments be mistaken for a sort of wistful "remembrance of things past," I'd easily place <strong>Terence Tao</strong>'s writings on par with Courant's, for many of the same reasons: clear and concise without being terse, straightforward yet not oversimplified, and, most importantly, animated by a sort of — <em>je ne sais quoi</em> — whatever it is, it seems to involve, in roughly equal proportions: mastery of one's own craft, a genuine desire to pass it on, and the considerable expository skills required to actually do so.</p> <p>Finally, I can't help but mention <strong>Richard Feynman</strong> in this context, and to plug his <a href="http://nobelprize.org/nobel_prizes/physics/laureates/1965/feynman-lecture.html" rel="nofollow">Nobel lecture</a> in particular. While not a mathematician <em>per se</em>, Feynman surely ranks among the twentieth century's best examples of a "mathematical physicist" in the finest sense of the term, not merely <em>satisfied</em> by a purely mathematical "interpretation" of physical phenomena, but surprised, excited, and, dare I say, <em>delighted</em> by the prospect! Moreover, he was equally excited about mathematics in general, see, <em>e.g.,</em> the "algebra" chapter in the Feynman Lectures on Physics.</p> http://mathoverflow.net/questions/48248/what-do-named-tricks-share/50450#50450 Answer by jasomill for What do named "tricks" share? jasomill 2010-12-26T22:14:32Z 2010-12-26T22:14:32Z <p>While tricks have names because they wind up being associated with with some particular mathematician, tricks are tricks because something important goes on "behind the curtain."</p> <p>For instance, to prove $$(a_1 b_1 + \cdots + a_n b_n)^2 \leq ({a_1}^2 + \cdots + {a_n}^2) ({b_1}^2 + \cdots + {b_n}^2),$$ write \begin{align*} A &amp;= ({a_1}^2 + \cdots + {a_n}^2)\\ B &amp;= (a_1 b_1 + \cdots + a_n b_n)\\ C &amp;= ({b_1}^2 + \cdots + {b_n}^2), \end{align*} then we must show $$B^2 \leq AC.$$ Equality clearly holds when $A = 0$. Otherwise, since $\mathbb{R}$ has no negative squares, for all $x \in \mathbb{R}$, $$0 \leq (a_1 x - b_1)^2 + \cdots + (a_n x - b_n)^2.$$ Expanding the squares, $$0 \leq Ax^2 - 2Bx + C.$$</p> <blockquote> <p>The quadratic expression vanishes whenever $$x = \frac{B}{A} \pm \sqrt{\left(\frac{B}{A}\right)^2 - \frac{C}{A}}.$$</p> </blockquote> <p>If $x = \dfrac{B}{A}$, then $$0 \leq A\left(\frac{B}{A}\right)^2 - 2 B\left(\frac{B}{A}\right) + C = \frac{B^2}{A} - 2 \frac{B^2}{A} + C = - \frac{B^2}{A} + C,$$ thus $$B^2 \leq AC.$$</p> http://mathoverflow.net/questions/22299/what-are-some-examples-of-colorful-language-in-serious-mathematics-papers/44992#44992 Answer by jasomill for What are some examples of colorful language in serious mathematics papers? jasomill 2010-11-05T20:52:43Z 2010-11-05T20:52:43Z <p>Spivak, <em>A Comprehensive Introduction to Differential Geometry, Volume 1,</em> p.94,</p> <blockquote> <p>Now that we have a well-defined bundle map $TM \to T\;'M$ (the union of all $\beta_x^{-1} \circ \alpha_x$), it is clearly an equivalence $e_M$. The proof that $e_N \circ f_* = f_\# \circ e_M$ is left as a masochistic exercise for the reader.</p> </blockquote> <p><em>Volume 3,</em> p. 103, indexed under "Idiot, any,"</p> <blockquote> <p>These normalizations are usually carried out with hardly a word of motivation, as if they are so natural that any idiot would immediately think of doing them—in reality, of course, the authors already knew what results they wanted, since they were simply reformulating a classical theory.</p> </blockquote> <p>From <em>Volume 5</em>, p.59,</p> <blockquote> <p>We are going to begin by deriving certain classical PDE's which describe important (somewhat idealized) physical situations. The word "derive" had better be taken with a hefty grain of salt, however. What I have really tried to do is give plausible reasons why the physical situations should be governed by those PDE's which the physicists have agreed upon. I've never really been able to understand which parts of the standard derivations are supposed to be obvious, which are mathematically simplifying assumptions, which steps are supposed to correspond to empirically discovered physical laws, or even what all the words are supposed to mean.</p> </blockquote> <p>Incidentally, Spivak gave an entertaining series of <a href="http://www.math.uga.edu/~shifrin/Spivak_physics.pdf" rel="nofollow">lectures on the subject of classical mechanics</a>, whence</p> <blockquote> <p>I haven't the slightest idea what any of this means! But I'm almost certain that it amounts to the similarity argument we have given. Aren't you glad that you aren't a mathematician of the 17th century!?</p> </blockquote> http://mathoverflow.net/questions/10535/ways-to-prove-the-fundamental-theorem-of-algebra/39298#39298 Answer by jasomill for Ways to prove the fundamental theorem of algebra jasomill 2010-09-19T11:52:17Z 2010-09-19T11:52:17Z <p>I'm partial to Milnor's proof in <em>Topology from the Differentiable Viewpoint</em>, a slightly simpler variant of the "every complex non-constant polynomial $p$ is surjective" proof given above, published somewhat earlier (1965). In brief:</p> <p><em>Definition:</em> Let $f: M \to N$, $M,N \subset \mathbb{R}^n$, $M$ compact, then, for each regular value $y \in N$, $$\#f^{-1}(y) = \text{number of points in the inverse image of y}.$$</p> <p><em>Lemma:</em> $\#f^{-1}$ is locally constant on the set of regular values.</p> <p><em>Proof of lemma:</em> Since $f$ is a diffeomorphism in a neighborhood of each $x_i \in f^{-1}(y)$, we can choose pairwise disjoint neighborhoods $U_i$ for the $x_i$, let $V_i = f(U_i)$, and then $$\#f^{-1}(V_1 \cap \cdots \cap V_k - f(M - U_1 - \cdots - U_k)) = \{\#f^{-1}(y)\}.$$</p> <p><em>Proof of the F.T.A.:</em> Using stereographic projection, we can consider the polynomial as a smooth map $f: S^2 \to S^2$. Since $f$ has only a finite number of critical points, the set of regular values is connected; the locally constant $\#f^{-1}$ is therefore constant on this set. As $\#f^{-1}$ cannot be zero for all regular values of $f$, it must be zero for none. Thus $f$ is surjective, and the polynomial has a root.</p> http://mathoverflow.net/questions/3559/colloquial-catchy-statements-encoding-serious-mathematics/62194#62194 Comment by jasomill jasomill 2011-05-10T10:48:39Z 2011-05-10T10:48:39Z &quot;X is the spitten image of Y&quot; is an informal phrase meaning &quot;X looks just like Y&quot;. &quot;X is the spitting image of Y&quot; is an entirely equivalent phrase; in some dialects, the pronunciation is even the same. As far as I know, neither is the &quot;right spelling&quot;; the phrase rarely appears in written English. http://mathoverflow.net/questions/123/linearity-of-the-inner-product-using-the-parallelogram-law/61908#61908 Comment by jasomill jasomill 2011-04-17T05:57:03Z 2011-04-17T05:57:03Z Very good point; I almost mentioned it, but I settled for writing &quot;n-dimensional&quot; instead. I can't imagine using this result as anything but motivation anyway, even in finite dimensions. Spivak mentions it while explaining why the Pythagorean theorem isn't <i>quite</i> &quot;true by definition,&quot; hence his choice of emphasis (that and it's a geometry book!). Pedantically: an ellipsoid E = {x in V: &lt;x,x&gt; &lt;= 1} contains a set B iff it contains the convex hull of B. Proof: sufficiency is clear, since B &lt; co(B). So assume B &lt; E, then co(B) &lt; co(E) = E, as E is already convex. (sorry, I couldn't resist...)