Should I call my former Analysis TA who's now an Assistant Prof. "Professor Last Name", when emailing them to touch base with something research-related?
user462942
We're on great terms, and he was a great TA to me. But I've only ever called him by his first name ... so it'd seem a little strange to call him Professor Last Name -- but I don't want to offend anyone either.
Nah, it's fine to call him as you always have. You can comment (humorously?) that you were tempted to call him Professor to celebrate his new status. :)
user462942
@TedShifrin Ok, cool -- thanks :)
user462942
@TedShifrin Can I ask you something?
user462942
I reread some of my writings -- journaling, sorta -- from about two years ago. I've taken quite a different direction since then for research. But I'm revisiting those ideas now, about doing abstract linear-algebraic research, such as functional analysis, operator theory, representation theory. Last night, I found an interview with a pretty famous prof. who remarked that Operator Theory was a dead field and that he urges his PhD students to enter into computational math / numerical analysis
user462942
12:15 AM
@TedShifrin do you think it's an accurate thing to say that Operator Theory is kinda dead?
@Astyx I don't know anything about the physics, but I've been told that the AQFT people also think a lot about operator algebras
But there's also people who apply operator algebras to group theory or rep theory by studying the group C*-algebras associated to a group, this has been generalized to semigroups, étale groupoids, etc.
Connes embedding conjecture was settled recently by quantum computing people
I'm forgetting everything. How do you prove that for a sufficiently nice $G$-representation on a vector space $V$, $k[V]^G$ is a finitely generated $k$-algebra?
Ah, alright. There's an operator $k[V] \to k[V]^G$ which mimics averaging that one can cook up which is surjective, and so $k[V]^G \cong k[V]/I$ for some ideal $I$
Hello people ! I have a vector space $\{V, +, *\}$ and I add to it an operation, $\circ$, such that $\circ$ is associative, distributes over $+$ and $(\alpha * x) \circ y = \alpha * (x \circ y) = x \circ (\alpha * y)$. Does someone knows the name of such a structure ?
My book is old school and calls this a "linear algebra"
then I think I read some proof like this: let $R$ be the subring generated by the coefficients of $\prod_{g \in G} (x - gt)$ for $t \in k[V]$, then $R$ is f.g. alg. and $k[V]^G$ is integral over $R$
I had an answer that I have been writing for about 6 weeks now, and had saved it as a draft in the "ask a question" tab. Today I found that my draft had disappeared. Is there any way to retrieve old question drafts?
Let $G$ be a compact group acting linearly on $\Bbb C^n$. Define $\Bbb C[x_1, \cdots, x_n] \to \Bbb C[x_1, \cdots, x_n]^G$, $f \mapsto \int_G g \cdot f d \mu$, where $\mu$ is the Haar measure. Is this well-defined?
For any ideal $I \subset R^G$, $IR \cap R^G = I$, since if an element of $R^G$ is an $R$-linear combo of elements of $I$, applying $\rho$ to both sides gives that the same element is an $R^G$-linear combo of those elements of $I$ (since $\rho | R^G = \text{id}$ and $\rho$ is $R^G$-linear); this immediately gives that if $R$ is Noetherian so is $R^G$
If $I_1 \subset I_2 \subset \cdots$ is an increasing chain of ideals in $R^G$, $IR_k = IR_{k+1} = \cdots$ stabilizes eventually, then just intersect back with $R^G$
That's not enough for $R^G$ to be a finitely generated $k$-algebra if $R$ is as such
It's definitely not true but what's an example of a Noetherian $k$-algebra which is not a finitely generated $k$-algebra lol
I suppose the point is $k[x_1, \cdots, x_n]^G \subseteq k[x_1, \cdots, x_n]$ is graded
If $A \subseteq k[x_1, \cdots, x_n]$ is a graded subalgebra which is also a Noetherian ring, it should be a finitely generated $k$-algebra, look at the ideal $A_{\geq 1}$ which is finitely generated...
This is some induction fact, if $f$ is a non-constant invariant element, $f = x^2 g + xy h + y^2 k$ because $(x^2, xy, y^2)$ is the ideal of all positive graded invariant elements, and $g, h, k$ have degree strictly less than that of $f$.
Maybe it should be possible to read off the quantitative Hilbert's invariant theorem from here, let me think.
Is the argument in general supposed to be: "Take the ideal generated by invariant polynomials; it's finitely generated; in lowest grading it agrees with the invariant polynomials; induct upwards to see that it agrees with invariant polynomials in all gradings"
This seems like you're telling me that the generating set for $k[V]^G$ is going to be homogeneous
I'm feeling skeptical
For instance, if you get some $V$ and $V'$ for which you get homogeneous generators in degrees 2 and 4, then $V \oplus V'$ should have non-homogeneous generating set i think
Yeah that's what I think this should relate to the quantitative version, Noether proved that the generating set is of size at most $\binom{n + d}{d}$ where $d$ is degree of the largest homogeneous element or something
Fucking hell how did I get here, all I wanted to see was that if $G$ is a compact Lie group acting linearly on $\Bbb R^n$ and $\zeta_1, \cdots, \zeta_n$ are generators of $\Bbb R[x_1, \cdots, x_n]^G$ then $(\zeta_1, \cdots, \zeta_n) : C^\infty_0(\Bbb R^n) \to C^\infty_0(\Bbb R^n)^G$ is surjective
Hey @BalarkaSen ! If we have an integer $a$ and when we divide it $b$, we have the following relation $$a = bq +r$$ Where $q$ is the quotient and $r$ is the remainder. But how can we prove that $r$ will always lie between $[0,9]$
@Knight on the number line, multiples of $b$ form dots that are $b$ distance away from each other (for social distancing measures let's require $b$ > 1.5 metre), so wherever you are on the number line you can walk leftwards for less than $b$ steps to find a multiple of $b$
In optimization technique, if we use integer indicator variables in the objective function multiplied with some constant value which is sampled from a non-linear function, would the objective function remain linear?
Can we talk of negative remainders? For example if $a\lt b$ then, can we write $$ a = bq - r ~~~~~~~~~~~~~r\in [0, b-1]$$ given that everything is an integer.
A banquet has to take place tomorow in the morning. There are 1000 bottles of wine, and one only is poisoned. You have rats to test the bottles, but if a rat ingest poison now, it will die just a few hours before the banquet. What is the smallest number of rats that have to be used in order to be sure to be able to determine which bottle is poisoned?
I don't know if this puzzle will interest many people, but maybe I should wait and give a hint later if it is necessary. Otherwise, the number is really small :)
Ah, @Alessandro's hint is good. We encode the wines in binary and use one rat for each "decimal" place. Rat $k$ drinks from all the bottles with a $1$ in the $k$th slot.
induction and I think then we can suppose we have infinite time so we would split in two parts of 500 bottles, use a rat that drinks all bottles from the fist part and reiterate
If you had infinite time to determine the bottle, you will use one rat that will drink the 500 first bottles. If it dies you know the poisoned one is in that group, if not it is in the other one. Then you use another rat to drink in the first 250 bottles of the group of 500 bottles in which you know there is the poisoned one ... etc
We also have to use 11 rats (because is the integer just greater than log_2(1000). I think binaries is just a trick to make all that in one time. So we don't have to suppose there is a limitation of time anymore. So if there is n bottles, you split in two groups with approximately the same number of bottles and you use one rat to eliminate one of the group, it seems that you can't do better with one rat. And then induction.
@Nûr: Still, you haven't proved that there's not a supremely clever algorithm that's better. You've just shown a second algorithm is no better (granting extra time).
I have a question about terminology (or maybe about the name of a subfield)...
So to preface, there are sequences of calculations which I will, for lack of a better term, label as 'arithmetic'.
Think of a succession of assignments, like in a programming language function, that for a few inputs goes through some directed acyclic graph of computation, and outputs a single number.
But suppose you only have partial knowledge of some of the inputs, let's say a range. And you want to see what the range of the answer is, dependent on the operations This is traditionally called 'interval arithmetic' (because instead of operating on point values you're operating on a range, like maybe an error range.
This might come up in numerical computations for seeing what the error in the output is depending on the error in the input.
So the question is... what is the arithmetic called when the input isn't just some point with error range, but an actual probabilistic distribution?
I'm not saying automatically/symbolically compute the theoretical lmiting distribution of a computation on distributions (eg sum of two normals has mean = sum of means and variance = sum of squares of variances..
I'm saying input is some arbitrary distribution (but histogram known).
Is this what is termed 'probabilistic programming'?
It seemed obvious to me that there should be such an area of computation, but I just didn't know of a name for it or had come across such a thing explicitly written about as an area of study.
@EnjoysMath Well, Im also just interested in -computing- the sample distribution from instances (of course figuring out a symbolic calculus for given theoretical distributions is very interesting too).
