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Martin Sleziak
Martin Sleziak
6:49 PM
You're right that my second expression for $\Basis_{i}'$ is awry; perhaps the moral is not to use partial derivative symbols except with local coordinates. :) Will fix this shortly. — Andrew D. Hwang Mar 13 '16 at 18:00
How do you know that you can "Pick an orthonormal basis $(\Basis_{j})_{j=1}^{4}$ of $\Reals^{4}$ such that $\Basis_{4} \not\in \mathcal{C}$"? Couldn't it be true that $\mathcal{C}$ contains the sphere $S^3$? I am okay with the argument if you say "orthogonal" instead of "orthonormal", but I wonder if I'm missing something and the argument is correct as stated. — Matthew Kvalheim Sep 9 '16 at 5:24
@Sandy.Davidson: If $\kappa_{n}(s) \equiv 0$, the last equation in (1) implies $\Basis_{n}(s)$ is constant. Since $\Brak{\gamma'(s), \Basis_{n}(s)} = \Brak{\Basis_{1}(s), \Basis_{n}(s)} \equiv 0$ in general, the fundamental theorem of calculus gives the equation you asked about, namely$$\Brak{\gamma(s) - \gamma(0), \Basis_{n}} = \int_{0}^{s} \Brak{\gamma'(t), \Basis_{n}}\, dt = 0.$$ — Andrew D. Hwang Sep 25 '16 at 21:32
@Sandy.Davidson: The function$$\Brak{\gamma'(t), \Basis_{n}} = \frac{d}{dt} \Brak{\gamma(t), \Basis_{n}}$$depends only on one variable, $t$. It vanishes identically by (1), so its integral vanishes; the integral has the stated value by the ordinary one-variable fundamental theorem of calculus. The unit vector $\Basis_{n}$ is normal to the hyperplane containing the image of $\gamma$. (Answering here instead of under your question so the comments don't get separated from their context. :) — Andrew D. Hwang Sep 28 '16 at 21:30
Do you mean you want to know in detail how to create a parametric curve to trace a conic in a particular plane? (The short answer is: Each of $\Vec{x}_0$, $\Basis_1$, $\Basis_2$ is an ordered triple of real numbers, so the given formula for $\Vec{x}(t)$ can be expanded to an ordered triple of functions of $t$.) — Andrew D. Hwang Oct 18 '16 at 13:49
Here, "translation" by $\Basis_{3} \leftrightarrow i(1, \alpha)$ amounts to rotation of a cylinder about its axis, while "translation" by $\Basis_{1} + \alpha\Basis_{2} \leftrightarrow (1, \alpha)$ amounts to translation of a cylinder along its axis. — Andrew D. Hwang Jan 11 '17 at 12:31
1. The tensor product of a vector and a covector may be viewed as a vector-valued linear function of one vector variable. In the second bullet point, $$T(v) = \sum_{i,j=1}^{n} a_{i}^{j} \Basis_{j} \otimes \Basis^{i}(v) = \sum_{i,j=1}^{n} a_{i}^{j} v^{i} \Basis_{j}.$$2. The sum you mention is the $n \times n$ matrix whose entries are all $1$. — Andrew D. Hwang Feb 12 '17 at 0:17
I tried the same queries with \basis - without any results.
I won't repeat the comment that contain also \Basis and have been posted above.
Doing the calculation explicitly in any dimension amounts to orthogonally diagonalizing a symmetric matrix; that is, it's hard to be explicit unless, say the matrix of $f$ is diagonal. I'm guessing you want to see how the inductive step works. The key point is, if $f$ is symmetric, and if $f(x) = \lambda x$ and $\Brak{x, y} = 0$, then$$\Brak{x, f(y)} = \Brak{f(x), y} = \lambda\Brak{x, y} = 0,$$i.e., $f(y) \perp x$. In other words, $f$ induces a symmetric operator on the hyperplane $x^{\perp}$. — Andrew D. Hwang Jul 26 '16 at 23:20
@Tanuj: The set of unit vectors is a hyperboloid, disconnected if $\Brak{\ ,\ }$ has signature $(1, n - 1)$ or $(n - 1, 1)$, and connected otherwise. — Andrew D. Hwang Aug 26 '16 at 21:41
@Tanuj: Small correction of earlier brain lapse: The set of unit vectors is disconnected if and only if $\Brak{\ ,\ }$ has signature $(1, n - 1)$, i.e., there is a one-dimensional subspace on which $\Brak{\ ,\ }$ is positive-definite, and an $(n - 1)$-dimensional subspace where it is negative-definite. (In the case of signature $(n - 1, 1)$, the "sphere" $Q(x) = -1$ has two components.) — Andrew D. Hwang Aug 27 '16 at 10:40
@theSongbird: I took the liberty of tweaking the math. (Less-than and greater-than signs space like operators; the delimiters \langle and \rangle are your friends here. :) Separately, are you sure the question didn't ask about $\Brak{A^{t}Av, v}$? — Andrew D. Hwang May 21 '17 at 10:39
Cool thing, bro, but do you have any idea as how to prove $\Brak{A^tA v, v} = \Brak{\lambda^2 v, v}$? — theSongbird May 21 '17 at 11:27
@Kolmin: You're welcome. :) Your picture of $\Reals^{\omega}$ as $\mathbf{N}\times\Reals$ is exactly the suggested "alternative", but do note that these are not the set; $\Reals^{\omega}$ is identified with the "space of sections" of the projection $\mathbf{N}\times\Reals \to \mathbf{N}$. — Andrew D. Hwang Jan 6 '15 at 20:17
@Odile: You're welcome. :) It's crucial to this argument that on a sphere, geodesics lie in a plane. Think of a helix on a circular cylinder in $\Reals^{3}$. — Andrew D. Hwang Jan 21 '15 at 17:45
The two definitions are equivalent, it's just that the composition is implicit in the definition with which you're familiar. :) When you write $$\phi(p) = \bigl(x^{1}(p), \dots, x^{n}(p)\bigr),$$you're thinking of $x^{j}$ as the $j$th coordinate function on $U$. That is, if $u^{j}$ denotes the $j$th coordinate function on $\Reals^{n}$, then $x^{j} = u^{j} \circ \phi$, exactly as in the "unfamiliar" definition. (By the way, welcome to MSE!) Edit: Yes, your edited comment is exactly right. — Andrew D. Hwang Mar 2 '15 at 13:25
It might be better to think of the $u^{j}$ as the natural consequence of the way a specific mapping is written as an ordered $n$-tuple of functions: If $\phi = (\phi^{1}, \dots, \phi^{n})$, then $\phi^{j} = u^{j} \circ \phi$. For your last question, "yes": In "$g(tu^{1}, \dots, tu^{n})$", the $u^{j}$ are the specific coordinates of $q$. Your book defines $G:[0, 1] \to \Reals$ by $G(t) = g(tu^{1}, \dots, tu^{n}) = g\circ \gamma(t)$, then uses the chain rule to compute $G'(t)$. :) — Andrew D. Hwang Mar 2 '15 at 14:23
Compactness comes from Heine-Borel: The sphere is a closed, bounded subset of $\Reals^{n}$. A covering by chart neighborhoods shows the sphere is a manifold (and since we know $S^{n-1}$ is compact, the covering shows $S^{n-1}$ is a compact manifold). :) — Andrew D. Hwang May 13 '15 at 10:19
If $M$ is a manifold, then every open subset of $M$ is a submanifold ("usually" non-compact). However, $M$ may also contain compact, boundaryless submanifolds of smaller dimension. The complement of a point in $S^{n-1}$ is diffeomorphic to $\Reals^{n-1}$, so every compact, boundaryless submanifold in $\Reals^{n-1}$ gives a compact, boundaryless submanifold in $S^{n-1}$. For instance, if $n \geq 2$, there are copies of $S^{n-2}$ inside $S^{n-1}$. — Andrew D. Hwang May 13 '15 at 11:05
Two more notes on terminology: 1. "Bounded" and "with boundary" have distinct meanings in this context. A subset $X$ of $\Reals^{n}$ is "bounded" if $X$ is contained in some ball, whether or not $X$ is a submanifold. "With boundary" unambiguously refers to a manifold-with-boundary, and usually suggests the boundary is non-empty. 2. If $X$ is a manifold, a "closed submanifold of $X$" is a submanifold that is closed in the topology of $X$ (logical); but to say "$X$ is a closed manifold" almost always means $X$ is compact and boundaryless (convenient, but vexing to the unwary). — Andrew D. Hwang May 13 '15 at 11:17
Finite sets of points are ($0$-dimensional) submanifolds, so even $S^{1} \subset \Reals^{2}$ has closed submanifolds without boundary. :) But perhaps this theorem partly represents your intuition: If $M \subset \Reals^{n}$ is a compact, connected $k$-manifold, then the only compact, open submanifolds of $M$ (i.e., boundaryless submanifolds of dimension $k$) are $M$ itself and the empty set. — Andrew D. Hwang May 14 '15 at 10:50
Two thoughts: 1. If we identify the unit ball in the $L^{1}$-norm with $[0, 1] \times \Reals$, we also have to identify the space of all continuous functions with $[0, 1] \times \Reals$. 2. If you consider measurable functions instead of continuous functions, the function space now has strictly larger cardinality than $[0, 1] \times \Reals$. So, I think you cannot truly identify these function spaces with subsets of the plane. (Sending a function to its graph maps a function space into the power set of $[0, 1] \times \Reals$, a much larger set.) :) — Andrew D. Hwang May 15 '15 at 13:49
Suffice it to say that every subset of $\Reals^{n}$ is Hausdorff and second-countable. :) Those properties are part of the "intrinsic" definition of a manifold mentioned just above the theorem, and make sense (only) in the axiomatic definition of a topological space. If you feel adventurous: Hausdorff and second-countable spaces. (To quote Morris Hirsch: "In a Hausdorff space, any two points can be housed orff [sic] from each other." — Andrew D. Hwang May 21 '15 at 11:52
@Yakk: Good point; I was implicitly assuming the knot is "tame", and a tame knot can be projected to have only finitely many crossings. The very first step (deform to a smoothly-embedded knot) breaks down if the original knot is wild. Offhand I don't know whether every wild knot in $\Reals^{3} \subset \Reals^{4}$ can be isotoped to an unknot. — Andrew D. Hwang Sep 8 '15 at 15:04
@Marc: The idea was to retract the open arc in $\Reals^{3}$ not keeping the endpoints fixed, i.e., to "untie the knot" while allowing the "lifted" interval to "ride along", held out of the way along the $4$th axis. :) — Andrew D. Hwang Sep 8 '15 at 15:35
Worse: It means there are no minimal regular heptagons in $\Reals^{3}$ (if "heptagon" refers to a hyperbolic polygon). — Andrew D. Hwang Oct 10 '15 at 17:17
$\Delta y = df_{a}(\Delta x) = f'(a)\, \Delta x.$. When you write $\Delta x.$ is that a value from the $T_{f(a)} \Reals$ space? — Brofessor Nov 10 '15 at 3:28
In this setting $\Delta x$ is a coordinate in $T_{a} \Reals$, which gets acted on (i.e., multiplied by) by $df_{a}$, yielding the element $\Delta y$ of $T_{f(a)} \Reals$. — Andrew D. Hwang Nov 10 '15 at 11:24
If $P_{1}$, ..., $P_{k}$ are mutually orthogonal $2$-planes in $\Reals^{n}$, and if a vector $\vx$ is orthogonally decomposed as $\vx_{0} + \vx_{1} + \dots + \vx_{k}$, with $\vx_{i} \in P_{i}$ for $1 \leq i \leq k$, an entirely analogous argument shows that rotating $P_{i}$ through an angle $0 < \alpha_{i} < \pi$ moves $\vx$ by $\sqrt{\sum_{i} \alpha_{i}^{2} \|\vx_{i}\|^{2}}$. — Andrew D. Hwang Nov 28 '15 at 12:55
If you only require $\phi$ to be an isometry along $S$, I believe you get a "tangentially-isometric slice chart" if and only if $S$ is (intrinsically) flat. (Flatness of $S$ is necessary, since $\phi^{-1}$ maps a small piece of $\Reals^{k}$ isometrically to $S \cap U$. I haven't carefully checked sufficiency, however.) — Andrew D. Hwang Jan 19 '16 at 21:35
Sort of: A derivation maps functions (or germs, depending on your viewpoint) to numbers; functions near $0$ are "naturally identified" with functions near $p$. To "differentiate" a function $f_{p}$ defined near $p$, identify it with its "avatar" $f_{0}$ (defined near $0$), and apply a derivation at $0$. This defines a mapping $l_{p}:\mathcal{F}(U_{p}) \to \Reals$ that satisfies the conditions of a derivation. (Details left to you. :) — Andrew D. Hwang Apr 4 '16 at 19:47
My thought was: If $v \in \Reals^{n+1}$ is the vertical projection (from $\Reals^{n+2}$) of a tangent vector to $M$, the orthogonal decomposition$$v = \frac{v \cdot x}{|x|^{2}} x + \left(v - \frac{v \cdot x}{|x|^{2}} x\right)$$gives$$D\phi(x)(v) = \frac{v \cdot x}{|x|^{2}(1 - |x|)^{3/2}} x + \frac{1}{(1 - |x|)^{1/2}} \left(v - \frac{v \cdot x}{|x|^{2}} x\right).$$This describes the image of $T_{x}M$ under $D\phi$, which fact might be enough (depending on the details of your situation) to calculate the normal field to the image without the generalized cross product. — Andrew D. Hwang May 14 '16 at 11:15
Yes, you're right: On geometric grounds, they're the minimum and maximum cosines of angles between $P$ and $P'$. These two planes span a space of dimension at most four, and without loss of generality, $P'$ is the copy of $\Reals^{2}$ in the first two coordinates. The unit circle in $P$ projects to an ellipse in $P'$ (possibly degenerate), whose semi-axes are these cosines. The product of the semi-axes is the scale factor for area. — Andrew D. Hwang May 19 '16 at 14:27
@studiosus: Yes, that was sloppy of me. You're thinking of, e.g., taking an arc-length reparametrization of $t \mapsto (t, \arctan t)$ and wrapping it around a product of unit circles, so the ends of $\Reals$ accumulate on a pair of parallel circles? — Andrew D. Hwang May 19 '16 at 22:00
I would read $h((x, y))$ as "one thing" and $f(x, y)$ as "two things". Again, some computer languages draw and enforce such a distinction to allow the compiler to manage memory accordingly. (A "pair" can be guaranteed to be stored contiguously; two numbers need not be.) That said, in Halmos' treatment, $\Reals^{2}$ and $\Reals \times \Reals$ are not just naturally identified, but actually identical. Ultimately, it looks to me that according to both Halmos and C++, "$f((x, y)) = f(x, y)$" is an abuse of notation, but for slightly different reasons. To Halmos, the abuse is harmless. — Andrew D. Hwang Jun 3 '16 at 14:27
By "doesn't appear to induce an isomorphism", I meant that the mapping $N \mapsto N+1$ induces a (bijective) mapping on sequences of exponents, but this is not the restriction of a linear operator. Note, incidentally, that mapping each prime to the next also does not induce an isomorphism, because $2$ (or, its coordinate axis) isn't in the image. Finally, each sequence of exponents may be viewed as real sequence, hence an element of $\Reals^{\infty}$. — Andrew D. Hwang Aug 7 '16 at 21:03
In the "current picture", the set of positive integers is identified with the set of sequences of non-negative integers with only finitely many non-zero terms. In that sense, there's not exactly a "$2$-axis", just a "first coordinate axis", which corresponds to the prime $2$ via (1). (Separately, and to re-emphasize, even if you construct an automorphism of $\Reals^{\infty}$, you may (or may not) have trouble defining a determinant.) — Andrew D. Hwang Aug 8 '16 at 10:49
How do you know that you can "Pick an orthonormal basis $(\Basis_{j})_{j=1}^{4}$ of $\Reals^{4}$ such that $\Basis_{4} \not\in \mathcal{C}$"? Couldn't it be true that $\mathcal{C}$ contains the sphere $S^3$? I am okay with the argument if you say "orthogonal" instead of "orthonormal", but I wonder if I'm missing something and the argument is correct as stated. — Matthew Kvalheim Sep 9 '16 at 5:24
@MatthewKvalheim: Let $S$ be an arbitrary $3$-sphere in $\Reals^{4}$. For fixed $P$ and $Q$, there are at most two real $t$ such that $(1 - t)P + tQ \in S$. The set $\mathcal{C} \cap S$ of points in $S$ having the form $(1 - t)P + tQ$ is therefore the smooth image of (some subset of) two copies of $K \times K$, which is not all of $S$. — Andrew D. Hwang Sep 9 '16 at 19:18
My second paragraph is perhaps easier to implement. Stereographic projection $\Pi_{x}:S^{n} \setminus\{x\} \to x^{\perp}$ expresses the round metric as conformally-Euclidean, so the differential maps the unit sphere bundle of $S^{n} \setminus\{x\}$ to a trivial sphere bundle over $x^{\perp} \simeq \Reals^{n}$, which can be scaled to the unit sphere bundle. Restricting to a closed hemisphere trivializes the unit tangent bundle over a closed hemisphere. — Andrew D. Hwang Feb 21 '17 at 2:42
Yes: If the image of $f$ is contained in $\Reals^{2}$ ("the image does not blow up"), the composition $\pi^{-1} \circ f$ sends $\Reals^{3}$ to $S^{2}$. — Andrew D. Hwang Apr 7 '17 at 23:06
In case it's of interest, here's an animation loop of the Riemann surface rotating in $\Cpx^{2} \simeq \Reals^{4}$. Note how the motion does look like rotation, but the entire plane containing the parabola is fixed. — Andrew D. Hwang Apr 29 '17 at 23:02
I don't have a primer in mind to suggest, but: The "graph" of a two-valued function is analogous to an ordinary graph, the set of points $(z, u)$ in $\Cpx \times \Reals$ such that $u = f(z)$ for some $z$ (and with two values for each $z$, which I'm taking to be $u$ and $-u$). Loosely, the set of non-zero complex numbers is a gasket; for the square root, walking once around the gasket causes the square root to pass from one sign/branch to the other. By the intermediate value theorem, the height has to pass through $0$, which causes the graph to self-intersect. — Andrew D. Hwang Apr 30 '17 at 15:22
Equip $\Reals^{2m}$ with the Euclidean inner product $g$, and the complex structure $J$ defined by $J(x, y) = (-y, x)$ for $x$, $y$ in $\Reals^{m}$. If we identify $(x, y)$ with $x + iy$ in $\Cpx^{m}$, then $g$ is precisely the Hermitian structure $$\langle(x, y), (u, v)\rangle = x \cdot u + y \cdot v.$$ A linear transformation $T$ is orthogonal if and only if $g(Tx, Ty) = g(x, y)$ for all $x$, $y$, is complex-linear if and only if $T(Jx) = JT(x)$ for all $x$, and (since the Euclidean and Hermitian structures agree) is unitary if and only if $T$ is both orthogonal and complex-linear. — Andrew D. Hwang Aug 9 '17 at 16:37
 

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