A harmonic function $f : U \to \mathbb{R}$ where $U$ is an open subset of $\Bbb{R}^n$ is a twice continuously differentiable function over $U$ that satisfies Laplace's equation $\nabla^2 f = 0$.
What's the significance of $U$ being an open set, @BalarkaSen?
I'm guessing it's to do with differentiation of some kind, but the specifics are currently hazy to me.
I would not think there's anything to do with differentiability. Usually in calculus domains of your functions are open subsets of $\Bbb R^n$ because all open subsets of $\Bbb R^n$ are homeomorphic to $\Bbb R^n$, so you don't have to deal with limit points, etc while working.
Yep, I see what you mean. I'm not entirely sure why either. In all the books I've seen, only one has mentioned that the domain may not necessarily be open, whereas the others have only mentioned that it's open.
Is that like a boundary point, @BalarkaSen? (One that may be in either a set or it's complement or the boundary of both)
Ah I've not encountered that before. The math world definition mentions induces topologies which I've not a clue on so far. Is there an intuitive definition for a connected open set, @BalarkaSen?
I mean, I've seen partial differentiation and the Laplacian involves second partial derivatives which may be treated as differentiating single variable functions w.r.t their only variables.
@Balarka: So you're saying... it's only defined on an open set. I think that is a more fundamental statement than "You have trouble with limit points".
Yep, I've seen that definition! That leads me to the one text where the domain of a harmonic function was said to not necessarily be open, @MikeMiller. In such a case, how would one define a derivative where the neighbourhoods of the points on the boundary aren't totally enclosed in the domain?
@Khallil: Given a good type of subset ("Codimension 0 submanifold with boundary") you can define differentiability, and the derivative, as follows. A function is differentiable if, around each point, there is a small neighborhood on which you can extend the function such that the extension is differentiable. (This definition makes sense for any subset.)
Then the derivative is just the derivative of this extension. This is where you need the subset to be good - this needn't be uniquely defined unless your subset is good.
Anotjer way of saying the above condition is "Closure of an open subset such that the boundary is a smooth, codimension 1, submanifold." If you're interested in this condition you can ask Balarka about it.
In practice the most important such subset when you're working with the Laplacian is the closed ball. In particular, the closed unit disc in $\Bbb R^2$.
Ok, that definition is coherent but perhaps a bit silly. Nobody cares about such a thing.
Oh sorry! I meant so do people only care about defining harmonic functions on certain domains to suit a particular problem (e.g. a PDE) that they're working on, @MikeMiller? (Not a very important question.)
@BalarkaSen, could you tell me more about codimensionality and submanifolds in an intuitive sense?
@Khallil: I would be very surprised if it wasn't a domain that was something like a manifold. Occasionally they allow looser conditions on the boundary; Lipschitz instead of smooth.
I recall the word smooth being thrown around a lot without much definition. I presume it involves infinite differentiability so that there aren't any kinks here and there, @MikeMiller?
@Khallil Loosely speaking, manifolds are subsets of $\Bbb R^n$ which are locally graphs of smooth functions.
So, $X \subset \Bbb R^n$ is a $k$ dimensional manifold if for any $x \in X$, there is a ball $B$ around $x$ such that $B \cap X$ is graph of a smooth function $\vec{f} : V \subset \Bbb R^k \to \Bbb R^{n-k}$
They are nice subsets of $X$, in the sense that you can't find local pathologies (they are locally just like $\Bbb R^k$!).
Codimension $m$ means dimension $n - m$. So codimension $1$ manifolds are dimension $n - 1$, 1 dimension lower than the euclidean space it lives in.
That's wicked stuff. So you locally intersect a subset of $\Bbb{R}^n$ with a ball at any point and you'll get the graph of a smooth function. Did you mean nice subsets of $\Bbb{R}^n$ as opposed to $X$, @BalarkaSen?
Koch snowflake is extremely pathological, in the sense that it's nowhere differentiable, let alone smooth. Smooth means it cannot even have a single singularity.
@Khallil Yes. And oops, yes, I meant nice subsets of R^n.
(there is a notion of submanifold of a manifold: $N$ a subset of manifold $M$ is a submanifold if $N$ is "naturally" a manifold, in a sense that the manifold structure is induced from $M$).
@Balarka: Well, what does "subsets which are manifolds" mean? The Koch snowflake can be given the atructhre of a smooth manifold, given that it's homeomorphic to a circle.
@MikeMiller I didn't say subsets which are manifolds. I said subsets which are manifolds in a natural way. To give a rigorous defn one needs to know the chart-language, I just didn't do it rigoriusly.
(as Khallil only asked for an intuitive definition)
Dogpiling. And I only do it to make sure you understand things correctly and that whoever you're talking to does too. Sometimes everyone understood; sometimes not.
I know, I appreciate it (e.g., I had never thought why the proof that every f.p group is a 4-fold group doesn't work for dimension 3). I was simply joking.
I think I figured out how to do the problem I was trying to do, but I am not looking forward to the actual computation of it…luckily, I think Wolfram Alpha can help
It's not too complicated but for a high school student like myself who hasn't even taken a calc course yet, integrating piecewise function with a bunch of absolute value stuff thrown in sounds a bit tedious since it has to be split into so many pieces
@RudytheReindeer A rather interesting application of Baire category theorem is that there is no continuous bijection $\Bbb R^n \to \Bbb R^m$ for $m > n$. In particular, space-filling curves cannot be injective. I don't see this mentioned in the answers oddly.
Sure. Assume $f : \Bbb R^n \to \Bbb R^m$ is some continuous bijection.
Then $f : B^n \to \Bbb R^m$ restricted to some ball $B^n$ (for some interval $I$ in $\Bbb R$) is a map between compact Hausdorff spaces, thus is an embedding. $f(B^n)$ is homeo to $B^n$.
$f(B^n)$ has to be nowhere dense, as if it was not, then it'd contain an $m$ dimensional ball inside, which is not possible. Now cover $\Bbb R^m$ by images of balls by $f$.
But BCT says countable union of nowhere dense sets has empty interior. This is impossible.
Hrm, actually, I think I need an argument for showing that an m-ball cannot fit in an n-ball.
@JulianRachman Well, surjective continuous maps $\Bbb R \to \Bbb R^2$ are precisely the space-filling curves, by definition. I am essentially just asking you to construct a space-filling curve.
The domain of $C$ would correspond surjectively to the interval $[0,1]$. However if we take it to $C\to[0,1]^2$, we may transform the function into $C\to[-1,1]^2$ by transformation and scaling. If we take the action between two cantor sets, it would be within $[-1,1]$. That is as far as I go.
Yeah, not sure if I understand what you're trying to say. But how are you modifying the surjective map $C \to [0, 1]$ (assuming we have found one) to a surjective $C \to [0, 1]^2$?
That's a good idea, but you need to provide the correct construction.
Suggestion: don't ponder over too many things at once. First, write down a surjective map $C \to [0, 1]$. Get this done, and then move on to $C \to [0, 1]^2$.
Don't expect that you'd finish it in a day. It took me a lot of time to figure out what could be the continuous surjective map $C \to [0, 1]$ possibly be.
Just keep it in your head, ponder on you free time.
(as a small hint, do not forget how we construct the cantor set)
@Julian: Just so you know, I gave you this exercise to make the point that concrete stuff can be fun. Homotopy theory maybe an important tool, but there are a lot of less abstract math out there which are equally fun. Space-filling curves give rise to a lot of geometry & topology - they were a part of Thurston's program.
@Balarka I fully understand and I thank you for exposing me. I will think about it and will give it my all. And as a side note, would algebraic geometry be a good subject for me to have a little more of a concrete experience?
While still connecting it to other abstract things that I am interested in?
No, it's as much abstract as homotopy theory is, depending on what "algebraic geometry" means (varieties or schemes?). And it requires a lot of background too, or so I have heard.
If you are algebra-minded, but enjoy some geometry (i.e., visual thinking) and want to see category theory lurking behind the scene, I'd say Galois theory.
@Stan: Not sure what you mean by "what does it do to vector fields". At every point, R(v,w) is a linear transformation, and that's what it does to the vector field at that point.
@Balarka: Abstract as the tools may be, to many algebraic geometers, the goals were quite concrete.
I wonder how successful schemes would have been if Grothendieck didn't have a Serre and Deligne and Dieudonne to ground him.
@Julian And Galois theory is a big branch. After you have learnt the basics, there is an immense amount of math you can learn from pursuing it (algebraic topology/algebraic number theory).
@JulianRachman Hm. Maybe Dummit-Foote, but it's not as good.
Artin has a chapter on algebraic geometry too. Admittedly it's not too much, but he talks about Riemann surfaces and uses it to motivate some Galois/field theory.
(and I second Reid, but I think one should know some algebra beforehand)
@MikeMiller Uh.....as I understand it, the Riemann curvature tensor measures the noncommutativity of two vector fields. So if I take R(u,v), this measures the non commutativity of these two vector fields u and v. But the Riemann tensor is rank 4, so it requires 4 inputs correct? I am having trouble underatanding what the additional two inputs are for.
If you ask me, I think Galois theory is worth spending the time than algebraic geometry. One reason is that it's more fundamental, but another is that Galois theory is broader than algebraic geometry. It can be the entry point of topology, number theory or algebraic geometry.
@JulianRachman The reason Reid is good is not that it's quick, but that it has a lot of concrete examples and geometric motivation (and of course, a lot of exercises).
@Stan: Well, it takes two inputs, really, and spits out a linear transformation $T_x \to T_x$. This linear transformation is "the other two parts" - one the input, one the output.
Excellent. I would recommend that without even looking at it, if it's Reid's.
I think complementing Artin with that lecture note of Reid would be the idea way to learn Galois theory. You can definitely try that. I just had a look, and the note's very concrete.
@MikeMiller in what sense is the linear transformation the other two parts? You mean if I take a vector then $w$, it maps it to $R(u,v)w$ and those are the "other two parts"?
@Stan: Yes. When we talk about the rank of a tensor, remember that tensors come in two parts. Some people call them covariant and contravariant, but I like to call them vector-like and form-like. An (m,n)-tensor takes m vectors as input and spits out n vectors as output. It has rank m+n - we add up the number of parts of the tensor of all kinds, vector-like and form-like.
The curvature tensor is a (3,1) tensor. It takes 3 vectors as input and spits out one vector. The first two input vectors are u and v. When we put those in, we now have an endomorphism of each tangent space - a (1,1)-tensor. That is, we plug in w, and we get the last bit, R(u,v)w.
When you (ughhhhhh) write this in index notation it's $R^{a}_{bcd}$. What this means is that if $\{e_i}$ is a basis of the tangent space, this is the coefficient of $e_a$ in $R(e_b,e_c)e_d$.
The fact that I have to sat "The coefficient of $e_a$ in..." is what gives it the right to increase the rank by 1.
@Stan: The one negative to this name (which I don't think is standard since I made it up) is that you have to remember - while forms are skew-commutative in their inputs, general tensors are not. If $T$ is a (2,0)-tensor there's no reason the number $T(u,v)$ should have much anything to do with $T(v,u)$.
I say "Form-like" because each index acts like a 1-form - it eats vectors.
Yeah. rank is the total number of vectors involved. Three as input, one as output. A (3,2) tensor (roughly) has 3 vectors as input and 2 as output so has rank 5.
(Really it has an element of $T_x \otimes T_x$ as output which is a finite sum of tensor products of vectors but whatever.)
In practice I did not like that class, learned some Riemannian on my own outside of it, and picked up a lot more while reading stuff either about or using Riemannian geometry.
I am not the model of how to learn it. (Also, I wouldn't say its part of any mathematician's toolbox. Just those who work near geometry.)
Well, I was able to find and use the method I suspected existed, unfortunately the ratio of time to accuracy is bad so it's not really worthwhile…I guess knowing what doesn't work is progress, eh?
So I have to compute some definite integrals, and finding the indefinite integral is very simple, but unfortunately it seems like I'm going to have to split it into around 30 intervals for the indefinite part…that sounds like fun
I ought to figure out how to do it in Excel so I can just plug in numbers
Given a $G$-bundle (usually $G=SU(n)$; sometimes $G=U(n)$, sometimes $G=SO(n)$; to me usually U(2), SU(2), of SO(3)) on an n-manifold, you can define a connection A on it.
Then $A$ defines a covariant derivative $d_A$. You can take its adjoint $d_A^*$. The Yang-Mills equation is $d_A^* F(A) = 0$, where $F(A)$ is the curvature of $A$.
This is the same as saying that $A$ has harmonic curvature, whatever that means.
We needed the metric when I took that adjoint.
@Balarka: I see, so because we never talk about calculus in here, you never learned it...
@MikeMiller I dont know a lot about adjoints. Why do we take the adjoint of the covariant derivative. What useful properties does that give us compared with the original $d_A$
Also, have u heard of the book Exterior Analysis by Suhubi?
I got a copy over the summer and it's a really nice book! Best intro book to the subject ive ever seen
"Any increasing (or decreasing) function certainly has an inverse, even if we are unable to give it explicitly (e.g., what is the inverse of the function $f(x) = x^5 + x + 1$?)." I think that $+x$ should have been $-x$. Not a big typo, if it really was, though.
@Stan: Forget $A$. The Laplacian on functions is $d^*d$, where $d$ is the exterior derivative. In general we define a Laplacian on forms by $dd^*+d^*d$. A form is harmonic iff $d\omega = 0$ and $d^*\omega = 0$.
If you're convinced it's natural to ask whether something is harmonic, you're convinced $d^*$ - and especially its kernel - is interesting.
Of note wrt the above discussion is that $d_A F(A)=0$ is one of the Bianchi identities. So all we have left to check to see if something has harmonic curvature is $d_A^*F(A)=0$.
Hahahah probably be worth it. Cheaper than the amount of time I spent lmao
Opportunity costs
We want to think of a geodesic as a curve in $M$ that is "as straight as possible." An intuitively plausible way to measure straightenness is to compute the Euclidean acceleration $\ddot{\gamma}(t)$ as usual, and orthogonally project it onto the tangent space $T_{\gamma(t)}M$
@MikeMiller I'm not really following how this helps me decide if a path is as straight as possible.
I like this french book, don't remember the names of the authors. Hulin, Lafontaine, ...? But it's in the Springer book dump.
@Stan: I want a number, not a vector in $\Bbb R^n$. If I don't, and my curve is in a submanifold, how can I even make sense of it as inherently something about the manifold?
Did you learn a little about curves in 3-space in calculus?
I guess it is true that if $f : U \times V \subset \Bbb R^{n+k} \to \Bbb R^n$ is map such that $f(-, \vec{y}) : U \to \Bbb R^n$ is a contraction mapping for every $\vec{y} \in V$, then the map $g : V \subset \Bbb R^k \to \Bbb R^n$ which sends $\vec{y}$ to the unique fixed point of $f(-, \vec{y})$ is continuous?
@Stan: That last explanation is a little weak because you caught me in a spot where my understanding is a little weak, but strong enough that I can't just say "Iunno".
