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8h
comment Adjoint representation is Lie algebra homomorphism
Is your first upper-case $G$ actually meant to be the Lie algebra $\mathfrak g$? And, usually upper-case Ad is $Ad(g):\mathfrak g\to \mathfrak g$ with $g$ in the Lie group. Then the derivative of Ad is lower-case ad, which seems to be what your question is about. Clarify? (At some level, the Jacobi identity is exactly the assertion that lowercase-ad is a Lie algebra hom, without reference to a Lie group.)
8h
revised Basic application of Weyl-Character-Formula
edited tags
10h
comment Can $\mathrm{PGL}_2$ be viewed as an affine algebraic group?
Linear groups modulo their centers have linear representations on automorphisms of their (algebraic) Lie algebras by (algebraic) Ad, generally.
1d
comment Norm map and extension of idele class groups
Ah, ok, ... I guess I don't have anything constructive to say about that particular item, sorry... maybe just thinking about the "class formations" and "Weil groups"... though that's still a bit in the opposite direction of what you suggest. Dunno...
1d
comment Norm map and extension of idele class groups
Okey-dokey, but isn't it more interesting to go in "the opposite" direction, and take a projective limit of $C_K/N_{L/K}$ over larger-and-larger $L$? Maybe I'm not understanding your goals here...
1d
comment Norm map and extension of idele class groups
Isn't that third term, $C_L/N_{L/K}$, simply too big, since $C_K/N_{L/K}$ is already the class group?
Jul
24
revised Question about the dirac $\delta$-function
edited body
Jul
24
answered Question about the dirac $\delta$-function
Jul
23
comment Functions with real domain but complex range, do they have any use?
I myself would vote on the side of Martin A., that in my experience the sanest default is that functions are complex-valued. One reason is that unitary operators are not diagonalizable on finite-dimensional spaces of real-valued functions... And characters of abelian groups really should be complex-valued, or else there'd be a lot of baggage about 2-D repns. Such stuff.
Jul
23
comment Show that a conformal map of an open disk onto any open disk is necessarily bilinear.
@WillJagy, :)....
Jul
22
comment Show that a conformal map of an open disk onto any open disk is necessarily bilinear.
@WillJagy, ... argh... Can't call a spade a spade any more? Wow. If I were younger and spunkier (and more optimistic that there would be change for the better) I'd rant about this... sigh...
Jul
22
comment Show that a conformal map of an open disk onto any open disk is necessarily bilinear.
@DavidC.Ullrich, indeed. :)
Jul
22
comment Show that a conformal map of an open disk onto any open disk is necessarily bilinear.
@DavidC.Ullrich, ah, sorry, I thought you were "defending", not "explaining". I suspect that the "bilinear" usage is just some "eggcorn" or whatever linguists call little semantic train-wrecks.
Jul
22
comment Show that a conformal map of an open disk onto any open disk is necessarily bilinear.
... and I don't think the associativity of the action, etc., is a bonus of calling non-linear maps linear. It's a bonus of the fact that the non-linear maps are descended from linear maps, on a quotient.
Jul
22
comment Show that a conformal map of an open disk onto any open disk is necessarily bilinear.
@DavidC.Ullrich, yes, indeed, and I say this to my students. But the linear fractional transformations are maps on a quotient, and are no longer literally linear. Sure, they're descended-to-the-quotient from linear maps, but... they simply aren't linear. If we decide to say that anything descended from a linear map, on a quotient, is called linear, then things will be rough, because we won't know any more whether it has the properties of a linear map... only that some one of its ancestors did. A nastily unhelpful terminological choice, I think.
Jul
22
comment Show that a conformal map of an open disk onto any open disk is necessarily bilinear.
@DavidC.Ullrich, things descended from linear things, that are not linear, are not necessarily ... linear. I don't know what you mean by "old", but nothing I read from late 19th through 20th in the professional literature called linear-fractional "linear". I've heard no person ever refer to them as such, in some decades of being in the math milieu. It is true that until maybe 1900 there was a not-so-clear sense of "linearity" in general, but, ...
Jul
22
comment Show that a conformal map of an open disk onto any open disk is necessarily bilinear.
I'd just like to note, for the record, that usually a "bilinear map" is a map $b:V\times W\to Z$ of vector spaces over a field $k$ which is linear in each argument separately. Seems to me un-necessary and unwise to subvert this very-standard, useful sense. Also, linear fractional transformations (also called "Mobius" transformations) were not ever called "linear", to my knowledge, in part because they are not literally linear ($f:V\to W$ of vector spaces satisfies $f(v+v')=f(v)+f(v')$, which few linear-fractional transformations do...)
Jul
22
comment Show that a conformal map of an open disk onto any open disk is necessarily bilinear.
@DanielFischer, the spate of references to linear-fractional transformations as "bilinear" seems to have kicked up pretty recently, though: I don't think I'd seen any such reference prior to a week or two ago (=early July 2015), and, since I am interested in some things related to such, I think I would have remembered any earlier episodes at all. But I see that you found some... Weird. Sorta obscures the meaning of "bilinear", for sure. What do they call bilinear maps, then? I don't understand this...
Jul
22
comment Show that a conformal map of an open disk onto any open disk is necessarily bilinear.
Rather than "bilinear", don't you mean "linear fractional" or "Mobius"? (And, indeed, as @DavidC.Ullrich comments, after you use a linear fractional transformation to assume wolog that the thing fixes the origin, Schwarz' Lemma implies it's linear (literally).)
Jul
21
comment Nature of the Continuum
The language of the question does remind me of A. Robinson's construction of non-standard reals, (to my perception) by modifying the equivalence relation one "mods out by" to make the ordinary reals (... non-principal ultrafilter...) "Taking standard part" (especially in E. Nelson's IST version) seems consonant with the tone of this question, etc.