@meow what do you know about about projective transformation? If you have $S,K$ subspaces of $V=\mathbb{P}^n(\mathbb{R})$ with $\dim S=\dim K=k$ is there always a projective transformation $\phi:V\to V$ such that $\phi(S)=K$? Why or why not?
that pdf on the other hand seems very accessible, I added it to the list of things I should read, which means I'll remember to do so in a few months maybe...
@Alessandro it's really cool doing it for GL(2,C) and classifying effects on the sphere under stereographic projection (see the youtube video "Mobius transformations revealed")
@arctictern I just like how visually intuitive the proof seems to be. Rotate $ge_1$ clockwise onto the x-axis, scale $ge_1$ by $1/r$ (which requires scaling $ge_2$ by $r$ to have det 1), and shear $r(ge_2)$ back to $(0,1)$.
If some user is interested in Analytic Number Theory they can see/study from the class of professor Ram Murty in YouTube from the official channel matsciencechannel, the name of the course is Introduction to Analytic Number Theory
Has anyone seen @robjohn recently? Has he transitioned into the grinch who stole Christmas, yet? Of do I need to wait a few days? Hmmm...I wonder how he'd look in green!
@MikeMiller So I think I just about proved that the Euler class is PD to generic zero locus. Now, do you know any nice examples of bundles with nonzero Euler class that are easy to provide using this? So far, I listed a bunch of non-zero Euler characteristic spaces, but I'd love to be able to use e.g. normal bundles for some examples.
I also don't have my notes available right now, if it doesn't bother you I'll ping you sometimes soon-ish so that you can look at my question when you have time, I'm not in a hurry
"if polynomials of degree n are equal on n+1 values then the polynomials are the same for all values" <- can someone say what i need to know to prove this?
If anyone has two seconds, this answer I posted here needs one more vote to reopen. Don't just cast a vote, but actually take a moment to look at it please.
I found the following formula in my previous question. This differs from my previous question in that I want an alternative proof of the below recursive formula for calculating $\displaystyle\sum_{k=1}^nk^p$.
Suppose I had a function recursively defined as
$$f(x,p)=a_px+p\int_0^xf(t,p-1)...
@MikeMiller which class you think I should take from the following three ? (1)Algebraic topology (2)Functional analysis 2 (3)Representation theory of lie groups.
@MikeMiller I was giving an example to Danu's self-intersection question. Didn't want to sound I was trying to contradict you; internet just keeps making bad timing.
@Adeek It depends on what you like the most. I like algebraic topology, I use functional analysis, I wish I knew the representation theory of Lie groups.
Locally convex spaces, weak topologies and duality in Banach spaces, weak compactness in Banach spaces, structure of classical Banach spaces, local structures, infinite-dimensional geometry of Banach spaces and applications.
In my last day on a complex analysis "set of informal lectures" with a person, he gave a smallish introduction to abstract Fourier theory. we started from measure theory and ended with infinite-dimensional representation theory of groups.
In mathematics, Faulhaber's formula, named after Johann Faulhaber, expresses the sum of the p-th powers of the first n positive integers
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k
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p
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But essentially one says that certain sets of particles that are physically related span a vector space on which the relevant Lie algebra acts (i.e. form a representation space)
Yes, and they can also be defined as the solution to some crazy contour integral (derived from that above), but I feel like I'm somewhere I don't yet belong is all. Don't worry too much, I'm sure I'll figure this out
@Sophie I was talking about the coefficients of the Taylor of your function ^. The coefficients of any Laurent series are representable by Cauchy's nth derivative formula
is there a fast way to check that $[Q(\sqrt(2),\sqrt(3),\sqrt(5)) : Q(\sqrt(2),\sqrt(3)) ] = 2$ without checking if $\sqrt(5) \in Q(\sqrt(2),\sqrt(3),\sqrt(5))$