@TedShifrin: Oh. If I write out the definition of $O(2)$, I get $a^2+b^2 = c^2+d^2 = 1$ and $ac+bd = 0$ for any matrix $$\begin{pmatrix} a&b\\c&d \end{pmatrix} \in O(2).$$
@TedShifrin: $a^2+b^2=1$, and $$\begin{pmatrix} a&b\\ -b& a \end{pmatrix}$$ and the same matrix but with $b$s having equal signs and $a$s different ones?
@MikeMiller Take the solid torus $X$. $A$ be the subspace obtained from taking the open trefoil knot and winding them around the hole of the torus, and pasting it.
@TedS been thinking about the aces. Can be done with hypergeomtric function, I -think- (haven't worked it out). But there is a nice way with indicators, too
I actually came to ask if anyone here knows a bit of prolog? I got an interesting question and I can't avoid the possibility of indefinite looping in it
I mean, it's a triviality that $S^3$ doesn't retract onto any knot. The retraction there is going to be wild in the sense that large points of $\mathbb R^3$ do not map to other large points.
@Balarka: I thought hard about a geometric solution and never could find one (although I'm sure plenty of geometric topologists could). But it follows from general nonsense because the open-ended trefoil knot is homeomorphic to $\Bbb R$.
I should be careful saying "general nonsense." It follows from Urysohn Lemma tools.
it amounts (i think) to me not really understanding the relation between the exterior powers of a vector space, and the exterior powers of the dual vector space
The composition part makes since, but why does it follow that the elements $x \in \Omega$ invariant by $x \mapsto x^q$ form a subfield $F_q$? I know that elements fixed by an automorphism make a subfield.
let me try to state this formally and see if i say it right. suppose I have an $N$-dimensional vector space $V$ with dual $V^*$. by taking antisymmetrized tensor products of $V^*$, i get the exterior powers $\Lambda^0 V,\Lambda^1 V,\cdots \Lambda^N V$
now, one thing that's first a bit perplexing. my usual (somewhat lame) way of thinking about dual vector spaces is to take the distinction between row and column vectors seriously
in that representation, elements of $V$ are column vectors and elements of $V^*$ are row vectors, and the latter acts the former to map vectors to $\mathbb{R}$ (or $\mathbb{C}$, in my case)
so, first question: in what sense are elements of $V$ functionals on $V^*$? is it just a matter of thinking about the operations going from left-to-right instead of right-to-left?
(i imagine the source of my confusion is my insistence on thinking of this in terms of row/column vectors, but i'd prefer to be doing things consistently)
i guess the way i'm thinking it is that, if i do insist on coordinate-dependence, then a column (row) vector acting on the right (left) of a row (column) vector is a perfectly sensible linear functional
@Semiclassic: In terms of your row/column vectors, it's symmetric to do $\mathbf v^\top\mathbf w$ as a linear map either on $\mathbf v$ or on $\mathbf w$.
Let's count dimensions for a minute. When $\dim V = n$, $\Lambda^k(V^*\otimes V)$ has dimension $\binom{n^2}k$. $\Lambda^k(V^*)\otimes \Lambda^k(V)$ has dimension $\left(\binom nk\right)^2$.
But you can contract elements in $\Lambda^k(V^*)\otimes \Lambda^k(V)$ by taking $(w_1^*\wedge \dots \wedge w_k^*)\otimes (v_1\wedge\dots\wedge v_k)$ to something like skewsymmetrizing the multiplications of the evaluations $w_i^*(v_j)$.
physically, it's about density operators. a density matrix equal to $(v_1^*\wedge v_2^*\cdots)(v_1 \wedge v_2 \cdots)$ should correspond to the system being in a pure state $(v_1^*\wedge v_2^*\cdots)$
taking linear combinations of such outer products gives you a quantum system with well-defined probabilities of being found in a given many-body state
the real question, which i'm still not quite grasping, is a claim regarding partial traces of density matrices of a composite system. in math language, i want to assume a splitting $V=A\oplus A^\perp$
and understand what $\Lambda^k V\otimes \Lambda^k V^*$ looks like in this case
which seems pretty painful :/
though i'm only interested in the subset of those wedge vectors which are totally decomposable
which actually reminds me of a much simpler question: how does one usually denote that subset?
@TedShifrin I wanted to ask you something ... I am reading centralizers and normalizers right now.. Can you tell me where they are used up in higher algebra... ?
I'm not thinking matricially here. But you naturally write $\bigoplus \Lambda^i(E)\otimes\Lambda^{k-i}(F)$, but you still haven't totally skew-symmetrized.
yeah. i should say, though, that the next step is to take a partial trace over $A^\perp$. that reduces the dimensionality to some extent, though not as much as i'd want
If you had a CPU with a finite number of bits, as much time as you wanted and a special way of making infinitely sized integers (like... 'start integer', byte byte byte... forever until the user decides to stop, then a 'stop integer') could the CPU add them together and finally print the actual integer? And why did that take me so long to write...? :D
@skillpatrol Thanks ! That's even more intriguing though why would it behave as the derivative of the acceleration :O it's supposed to exist even with a constant acceleration
@ParanoidPanda - If you don't like what he's saying, you can always set him to ignore
@wolfboyft - Paranoid clearly doesn't appreciate your attention. If you persist in contacting him/her/them, this is likely to lead to you taking a short break.
@dREaM - Please don't provoke matters.
user136984
@Richard: I wish to keep an eye on him just in case if he gives away more of my personal information as he did before, so I can't put him on ignore, but I will now ignore him, and just flag up anything he says which gives away my personal information.
@MikeMiller Just to clarify, I understand the Tietze argument can be applied to prove that R^3 retracts to the knotted x axis, but is it necessary that a similar argument can be applied to prove that solid torus thingy? I mean, the closed subspace A in that case is homeomorphic to S^1, and Tietze is about maps from space to R.
Well, for R^3 and the knotted x-axis A : $f : A \to \Bbb R$ be a homeomorphism. Extend to $\tilde{f} : X \to \Bbb R$ by Tietze. Compose with $f^{-1}$. The resulting map leaves $A$ fixed, thus is a retract.
I literally don't know of any tool which I could use to approach it. The two things I use for proving things don't retract into things is 1) injection of \pi_1 2) homology split exact sequence. Both are useless here.
Every ring is isomorphic to a subring of the endomorphism ring of it's underlying group. That's Cayley's theorem for rings.
What can we say about rings that are isomorphic to the endomorphism ring of their underlying additive group?
(are they always integral domains? ADDED: no, $Z/nZ$ is a cou...
@MikeM: The injectivity radius of $U(n) is well known (perhaps $\pi$). One would just have to figure out how many translates are needed to cover. All well-known, I'm sure, just not at the moment by me.
Every ring is isomorphic to a subring of the endomorphism ring of it's underlying group. That's Cayley's theorem for rings.
What can we say about rings that are isomorphic to the endomorphism ring of their underlying additive group?
(are they always integral domains? ADDED: no, $Z/nZ$ is a cou...
