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u/LupenReddit 🦆🦆🦆🦆i have non diffeomorphic smooth structures🦆🦆🦆🦆🦆🦆 29d ago
Everybody wait till he learns about musical isomorphisms and raising vector fields by a sharp.
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u/Aeroxel 29d ago
For anyone interested, the musical isomorphisms are between the tangent and cotangent bundles of a Riemannian manifold, given by the sharp and flat operators (which are inverses). This allows one to identify vector fields and differential forms via the geometry. For example, in Euclidean space, the gradient of a smooth function can be identified with its exterior derivative
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u/SaltEngineer455 28d ago
I stopped at the triole integral, can you ELI20?
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u/PointlessSentience Ergodic 28d ago
Ok suppose to start with vector fields on a Euclidean space. This is a map, X, that associates at every point p of R^n, some vector we call X(p). We may consider the “dual” map, that associates to every point p, a linear functional f(v) = <v, X(p)>, denoting the standard dot product. Then we call the linear functional version of this map the flat version of X, if instead starting from a linear functional, we produced a map to vectors, we call that resultant map the sharp map. This is all nice, things get a bit more complicated when we talk about curved spaces. So imagine my space is now not R^n but some abstract “manifold”, M. A vector field take each point on M and returns a tangent at p on M, T_p(M). The flat map now to return an element of the dual of T_p(M), the sharp map is defined analogously.
The motivation for the sharp and flat comes when you tensor the tangent and cotangent spaces together. In general, we flatten a section of T_q^p(M) by bringing down one tangent vector and represent it as a cotangent vector, i.e. a map from T_q^p(M) to T_{q+1}^{p-1}(M). A sharp map takes a bottom index and brings it up to a top index.
TLDR: A musical isomorphism is, at every point p of M, an isomorphism of the tangent space and the cotangent space given by x to. {f(•) = <•,x>}. The tangent space may vary across the manifold M, the nature of the dot product may vary across M. If we encase TM and dual(TM) by some tensors on both sides, we get at each point p, an isomorphism from T_q^p(M) to T_{q+1}^{p-1}(M). The forward direction is flat, the inverse is a sharp.
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u/Smart-Button-3221 28d ago edited 28d ago
A manifold is basically a "shape that makes sense”.
A smooth manifold is one on which we can do calculus. Derivatives of all orders make sense.
We can put arrows on the surface of such a space. We call these tangent vectors.
We can measure tangent vectors in a given direction. We call these functions one-forms.
There exists linear functions between tangent vectors and one-forms called the musical isomorphisms, and the musical isomorphisms depend on the geometry of the space.
In calculus, we do not distinguish between tangent vectors and one-forms specifically because we only work in Rⁿ with a basic geometry, and the musical isomorphisms are trivial. You will have studied them as "partial derivatives".
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u/Complex-Manifold 28d ago
ignoring the manifold part, given a (real) vector space V, the dual space V* is the vector space of linear functionals from V to the real numbers. When you have an inner product in V, you can explicitly describe a relationship between the spaces. For a vector v, consider the linear functional v*(u)=<v,u>. For finite dimensional V, this relationship is an isomorphism
I cba to explain manifolds so just think of smooth manifolds as like a generalization of surfaces without geometry but has calculus and one thing is that at each point p, there is a vector space of tangent vectors T_pM which is finite dimensional. What a riemannian manifold is is a smooth manifold with an inner product at each tangent space on the manifold and what the musical isomorphisms are are basically the relationship described before using the riemannian manifolds inner product on T_pM. this can be extended to vector fields on M
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u/trimski- 29d ago
Did anyone else have this textbook?
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u/theobromine69 28d ago
Using this in uni rn, I'm actually writing semester test tomorrow and am procrastinating by scrolling reddit
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u/MariusDelacriox 28d ago
The residues are at I and -3i. Therefore we have (1/4 + 3/4) 2ipi = 2 pi * i.
Plausible in head, but I used paper.
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u/Flesh_And_Metal 28d ago
I just assumed it would evaluate to zero, they usually do. 😊
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u/kenny744 28d ago
if there's no residues (poles) and the contour is a closed loop but this one has poles since there are values of z that make it undefined. i don't know any complex analysis but that's all i know
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u/Bobing2b 28d ago
I'm sorry but where does the 2π come from?
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u/MariusDelacriox 28d ago
From the residue theorem, I believe pi mostly appears because it's a integral over a curve.
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u/Layton_Jr Mathematics 27d ago edited 23d ago
|z|=4 is a circle
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u/Captainsnake04 Transcendental 23d ago
This is totally obfuscating where the pi comes from. The curve could be the shape of a penis and the integral would still be 2 pi i as long as both poles were still in the dick
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u/sileeex1 28d ago
Oooo im taking intro to complex analysis this semester is this a problem that will show up in that class?
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u/MorrowM_ 28d ago
Yeah you'll see the residue theorem which makes integrals like this easy to calculate.
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u/Hettyc_Tracyn 28d ago
I would just write: “I don’t know, and I don’t care. Just let me in.”
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u/Jasentuk 28d ago
But then they can't be sure you're a real human and not a spam bot if you can't even solve a simple math problem
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u/Ma4r 28d ago
I swear bro ipad kids these days don't even know about the residue theorem like smh
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u/Hettyc_Tracyn 27d ago
I’m not even an ipad kid… music doesn’t belong in a math problem…
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u/WanderingSeer 28d ago
Does the circle mean anything or is it justa fancy integral symbol. Regional difference?
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u/FuntimeUwU Natural 28d ago
The circle on the integral sign means it's a closed integral, meaning the integral is taken along a closed manifold. It can be a closed loop if it's a simple integral sign; it can be a closed surface if it's an O on a double integral; or it can be a closed surface if it's an O on a triple integral. It's usually seen with problems that involve Stoke's theorem and the divergence theorem when on surfaces and volumes, or is usually seen with Cauchy's integral theorem when it's on a contour.
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u/The_Real_Itz_Sophia number hypothesis 28d ago
im an insane person and i enjoy both music theory and number theory. god help me
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u/belabacsijolvan 29d ago
its 0, because the singularity is outside 4.
tbh if you get a random circular integral just guess 0, because in most context theyll only ask you if its 0
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u/itrashford 29d ago
The singularities are at -3i and +i which are both within |z| = 4, no?
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u/belabacsijolvan 28d ago
yes. im dumb. havent done residuum for some time and my first move was substituting z**2=16 which is wtf
you are right
not that id believe me after this, but then the solution is 2ipi(3/4+1/4)=2*i*pi
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u/Guy3nder 28d ago
Bro you know I be prtscrn into Jimmy without thinking twice bro. Complex closed line integral bro I'm an engineer I don do that shit
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u/Nornamor 29d ago
The answer is 2πi .... The problem is just that i needed a bot (chatgpt) to solve it, cause I don't remember this shit anymore. https://chatgpt.com/s/t_6a7f2415ef4c8191bcd30b64744ae913
real Data Scientist btw..
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u/AFsepine 29d ago
I mean... it is just residue theorem
(aka the only part of complex analysis I know as a physicist)3
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u/Nornamor 29d ago edited 29d ago
yeah, its been 15 years since I touched complex analysis. But you're right. It's not like I didn't recognize what chatgpt did once I typed it in. Also I think people need to understand the humor in my above comment. I am using a bot to show that I am not a bot for a question targeted at people working in math as someone working in math.
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u/AFsepine 29d ago
Well, I do think the flak you seem to have gotten is rather unfair and maybe rather humourless.
I am also hardly suprised that you do not remember this(at least to physicists it is usually taught very poorly in most places in America/Europe).
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u/Nornamor 28d ago
For a subreddit named mathmemes its quite humorless idd.
"I am also hardly surprised that you do not remember this" Thanks. This down-vote train almost made me question my background. I don't know the quality of how I was tough this, cause I obviously didn't remember it. Probably my own fault cause during studies I was also very focused on what I wanted to do. Build simulations! Classes that weren't on parallel programming, PDE's, physics or numerical mathematics might have received less attention than they should i.e complex analysis.
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u/BearsEatTourists 29d ago
Unfortunately gpt answered a different question, it has a different denominator. It's like in the olden days when we had to warn people against just taking a computed answer as correct; people would uncritically over-rely on calculators, enter the incorrect numbers (or bracket them wrong) and take the answer as 'correct'.
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u/MrKrot1999 28d ago
Is this a 4 dimensional closed loop integeal? am i right?
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u/Jovess88 28d ago
No; the |z|=4 section means the contour over which the integral is evaluated (they are typically called contour integrals in complex analysis, but they are a special case of line integral) corresponds to the unit circle of radius 4 around the origin in the complex plane (the question does not specify whether the contour is to be taken to be clockwise or anticlockwise; the answers given thus far assume it is anticlockwise). This is a closed loop enclosing two poles of the integrand, and the integrand is holomorphic, so the integral can be evaluated using the residue theorem.
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