Any reason why you didn't mention the idea of determinant telling us the factor by which linear maps distort signed volume? I had watched the determinant vids, but you didn't talk about it, but it is mentioned in the intro paragraph of the next section.
Maybe I could come back to it after I do section 6 and it will be a clearer idea then.
I'm not right now.....I usually do a general read of the section to get the concepts down, then read it a second time trying to do the proofs of theorems.....But COV will wait for another day when I'm done the book....
I was looking at the discussion of these concepts in the textbook my school uses and what I used when I first did it..........set up for failure...
nobody would be prepared to address these concepts based on the level of pre-reqs for the course.
Does anyone know of some papers which talk about rephrasing quantum theory in terms of the language of category theory and if this field of work is still active? I have found a number of papers around ~2010, but not so many of recently
I’d like to show that a Riemann surface $X$ is hyperelliptic iff there is a holomorphic involution $\tau : X \rightarrow X$ such that $\tau$ has exactly $2g-2$ fixed points, where $g$ is the genus of $X$.
(I’m working with the following definition of a hyperelliptic Riemann surface: $X$ is hypere...
koro: your recipe for "F" doesn't make sense if f is only given on the circle. note that cauchy's integral formula tells you what the analytic extension would have to be.
and e.g. the formula for laurent series coefficients tells you a bunch of integrals that will have to be 0, if there is to be such an extension.
maybe explore what can happen in a nonconstant example, e.g. f(z) = z for z in the upper half of the unit circle and f(z) = 1/z on the lower half of the unit circle.
and as a side note, you seem inordinately obsessed with this. there's a lot of nontrivial analytic function theory relating to taking stuff inside the disc/circle and extending it beyond the disc and vice versa, and if nobody is asking you to do this, it's OK to move on.
come back later, if at all. a lot of people do complex analysis in ten million different directions, and don't get into exotic function spaces, even on the circle or unit disc.
maybe. it wasn't clear from koro's initial question if he wants the extension to be analytic on the open unit disc only, or on all of the closed unit disc. there are a bunch of restrictions that immediately pop up if he meant the latter (e.g. relating to continuous functions not necessarily even being differentiable), so i assumed that he didn't. but yes.
"just write down [a formula] that extends it" works a ton of the time in specific examples, but certainly not in this case, with an arbitrary continuous function as input data
or an arbitrary continuous function on the upper half of the circle. the symmetry condition there isn't much of a restriction, for purposes of providing enough structure to make the analyticity work out.
Then gpt's usage of identity theorem is wrong. It assumes analytic on $\overline{\Bbb D}$. And also, derivation of contradiction is weird. It doesn't contradict anything
Though, it's not what I wanted. I know already (from our earlier discussions) that if f is real valued, then the extension is not possible if f is non constant on the boundary.
@onepotatotwopotato because it satisfies the boundary condition.
I was very confused earlier as you seemed to be suggesting otherwise and also the comments on my post.
That's probably I misstated the question as one of the comments to my post also said.
I used the words 'Schwarz's lemma' etc. for my question but they have a different meaning in complex analysis, i.e., "Schwarz's lemma'' is reserved for a result.
As Leslie mentioned earlier, the question is not clear as 'the extension' means holomorphic on the boundary also apart from the disk or only holo. within the disk.
I assume the former to apply Cauchy integral formula.
Is Cauchy's integral formula valid if f is holomorphic on closure D?
Can I choose the radius to be equal to 1?
(f is not even defined outside closure D.)
because the statement usually says: f is holo. on a domain containing closure of the unit disk.
what is the meaning of such f to be holo. on boundary of D?
the obvious meaning just as we have in case of left and right limits?
If the extension meant the latter, then I think that any hol. extension will do.
*any entire function
That is, suppose that f is holomorphic on closure D, then for any z in D, we have $f(z)=\int_{\partial D} f(u)/(u-z) du$. Is this true?
Looking at the proof of Cauchy's integral formula, I think that the answer is in affirmative.
A book says the curvature $K$ of a metric $\lambda(z)|dz|$ on a simply connected Riemann surface is given by $$K = -{\Delta\log\lambda\over\lambda^2}.$$ What is the curvature here? (e.g. sectional curvature)
I remember there are several curvatures in Riemannian geometry
Suppose that $f: \mathbb S^1\to \mathbb C$ is continuous such that $f(z)=f(\bar z)$ for all $z\in \mathbb S^1$. Can it be extended to a continuous $F: \overline{\mathbb D}\to \mathbb C$such that $ F$ is holomorphic on $\mathbb D$?
If $f$ is constant, then the result is obviously true.
So suppose ...
In the language of Dirichlet series and the Riemann zeta function I believe this could be counted as an elementary proof:
Add the variable $s$ as an exponent to your series so that it becomes:
$$\sum_{n=0}^\infty\left(\frac{1}{(6n+1)^s}+\frac{-1}{(6n+2)^s}+\frac{-2}{(6n+3)^s}+\frac{-1}{(6n+4)^s...
On the number of prime numbers less than a given quantity. <=> On the number of exponentiated negated Riemann zeta zeros divided by c, divided by pi, below a given constant c.
Can somebody tell me how should I go with solving the following problem:
We have a laplace equation $u_xx + u_yy = 0$ with boundaries $u(0,y) = \epsilon \sin(y/\epsilon)$, where $x$ lies in positive real numbers with zero and $y$ lies in all real numbers.
So as far as I read, should I just use the separation of variables technique ?
@PlaceReporter99 Ok. Have fun with that. ;) I think there's some mention of that sort of thing on the Wikipedia page I linked, but I know almost nothing about fractional calculus.
@PlaceReporter99 BTW, I've been investigating various approximate solutions for $0\le x\le1$ in $x/y=x^x$, given $0\le y\le1$.
I'm getting interested in non $C^\infty$ manifolds. Given a derivation on $C^k$ functions, why can't it act on a $C^h, 0<h<k$ germ? Being a 1st order operator, wouldn't it make it a $C^{h-1}$ function?
I ask that because I'm trying to understand why the tangent space construction as a space derivations fails
There are various ways to answer that. First, the space of derivations is infinite-dimensional. Second, the usual trick of writing a smooth function as $f(x)-f(0) = \sum x_i g_i(xj$ with $g_i$ smooth messes you up with $C^k$.
What should I be doing to get my numerator to be zero? I am mistating something but can't figure out what. I looked at a few answers on MSE, but I wanted to do it the "textbook" way
$T$ is my "candidate" for the linear map. So to be differentiable at a point $a$ means there is a linear map $Df(a)$. In this case $Df(a) = $(a)$, plus the rest of the difference quptient stuff
The full definition being a function $f$ is differentiable at $a$ if there is a linear map $Df(a): \mathbb{R}^n \to \mathbb{R}^m$: so that $\lim_h \to 0 \frac{f(a + h) - f(a) - Df(a)h}{\|h\|} = 0$
dc3: ted understands what you're talking about, but is checking to see if you understand what you're talking about :) i think the game he is playing is that if you can explain it to him, your confusion will vanish
I'll make a stupid example (in very plain language) to make clear what I don't understand about it. Consider $\partial/\partial x^i$ acting on $C^k$ germs, then I can also consider a "copy" $\partial/\partial x^i$ of it acting on $C^{k-1}$ germs. In this sense it is also a $C^{k-1}$ derivation. If the above is a problem, does that mean that there are derivations in $C^k$ such that there is no corresponding derivation in $C^{k-1}$?
@TedShifrin That is a problem because I end up with functions that aren't necessarily differentiable. As long as $k>1$, since derivations are first order operators, I don't see what fails
ted: it's a private pool that very few people use, so it functions as a pond. the ducks don't seem to mind us, sometimes they land in it while we're using it.
You don’t begin to understand what derivations look like when we’re not in the smooth world. That lemma is what allows us to understand the smooth case.
And I don't know how to do that (at least in general)
I think I have clarified the problem. I have been doing circular reasoning: thinking of those "simple" derivations $a^i\partial_i$ but what allowed me to do that in the inf. dim. was the lemma we talked about. I can't use this consequence of the lemma to prove the lemma (i.e. recklessly say that I can apply a $C^k$ derivation to a $C^{k-1}$ germ)
dc: the "(a)" in Df(a) is a label telling you what point you're considering Df at. of course Df(a) is also a linear map and hence takes a vector as an argument. but this aspect of the "(a)" notation isn't too compatible with actually writing out what a is when you're applying Df(a) to a given h
i don't think it's a good idea to signal the potential dependence of Df(a) on the point a by writing out the components of a in the middle of expression involving Df(a) or Df(a) applied to something else
so up above, i would scrub out the column vector "a" between the 3x2 matrix (presumably representing Df at the point you're fixing, which you might be calling a) and the vector h
just understand that for general f, the 3x2 matrix Df(a) has entries that depend on a
for example if f(x) = x^2 in calc 1, and i evaluate f' at for example at a = 3 i get that f'(3) = 2*3 = 6. but i wouldn't write this as 6(3), even though 6 definitely is what f' is when evaluated at 3, and even though in general, what f' is would depend on what point i'm evaluating it at
i don't need to expressly note the dependence on my choice of a by writing the value of a next to what the derivative is
it's a problem of juxtaposition being used to denote both "evaluation at" in general, and also, with linear maps, that's matrix multiplication with the thing you're evaluating at
in something like Df(a)(h), the juxtaposition of Df with (a) means "Df evaluated at a" [which is not, in general, a product of some matrix, on the one hand, and the column vector a, on the other - in particular, Df(a) has type "matrix" and not type "vector"], while the juxtaposition of Df(a) with (h) means both evaluation of the linear map Df(a) at h, and a matrix-vector multiplication of the matrix Df(a) with the column vector h
it looks like maybe sometimes your book up above is trying to keep this a little separate by not putting "evaluation at" in parentheses when doing matrix multiplication. they write Df(a)h and not Df(a)(h)
but i wouldn't count on other books to do the same
well at least I have a better comprehension of this now so when I do encounter it again in higher courses
@TedShifrin Right. It was me using $T$ and not thinking about it as a matrix in the derivative form that messed things up......but it was a good learning lesson
heh, i checked to see how rudin handles this, and he includes two sentences by way of exposition that he does not include for other complicated stuff.
those sentences below the equation display (19), haha. he never says stuff like that, but puts it there. basically for the reasons discussed above, dc3.
@PM2Ring could you help me plot the gram series in sage? I tried looking up documentation but couldn't figure out the double series aspect of this function. mathworld.wolfram.com/GramSeries.html
So I do remember that the derivative operator is a linear map.....so we are applying the "linear map" (derivative operator) onto an object which in this case is a linear map from the set of linear maps.
the n = m = 1 case is sort of illuminating because at least the calculations are clear, maybe but also sort of confusing because the mental scaffolding around the idea is different. it's maybe not very calc 1 to think of numbers both as numbers and as linear maps on a one-dimensional vector space (6 vs. "multiplication by 6"). also thinking of the thing that plays the role of "h" as an argument is maybe not very calc 1, depending on how/whether you cover differentials.
but at the galaxy brain level this is no different from thinking of a matrix as a matrix and also as a linear map
In the language of Dirichlet series and the Riemann zeta function I believe this could be counted as an elementary proof:
Add the variable $s$ as an exponent to your series so that it becomes:
$$\sum_{n=0}^\infty\left(\frac{1}{(6n+1)^s}+\frac{-1}{(6n+2)^s}+\frac{-2}{(6n+3)^s}+\frac{-1}{(6n+4)^s...
