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1:59 AM
I am not sure why this is off topic? I am not asking a specific homework question, I think? I changed the numbers of the homework question. I just don’t know how to rearrange the formula to get this to work.
4 hours later…
5:56 AM
@JohnRennie Got a question about a solenoid with an iron core and without. (Aircore) I am wanting to produce an enormous magnetic field strength. What is it actually that creates magnetic field strength and makes it stronger? Is it the current or is some sort of current and voltage ratio? I know I can't have high voltage and large field strength. And I know I can't have thousands of amps and have 10 volts and generate a super strong magnetic field strength.
Sorry for the short length and poor abbreviations. Wish I could send a longer message with all details. But stack exchange has a character limit. I am looking to make an extremely powerful electromagnet and I need to sort it out once and for all. Which one actually gives a magnetic field more strength, current or voltage or a ratio of current and voltage?
Anyone else that would like to chime in. Please do, wouldn't be a bad idea to get everyone's input.
2 hours later…
8:25 AM
the magnetic field comes from the current
and the current is determined by the resistance and the voltage
so you can see that eg. voltage choices can effect the magnetic field, but it is by the effect of voltage on the current that it does so
2 hours later…
10:03 AM
@TheDragonOfFlame Our homework-and-exercises policy doesn't care about whether or not this is an actual homework question (because how could we possibly tell?), but whether it has the form of one. Asking just how to solve an exercises or perform a specific computation is off-topic regardless of the origin of the question
10:44 AM
mostly related to [this question](https://physics.stackexchange.com/q/668175/58382), though from a slightly different perspective: in physics literature, people usually have the covariant derivative seemingly act on components of fields. Eg one writes things such as $\nabla_\mu V^\nu = \partial_\mu V^\nu + \Gamma^\nu_{~\mu\eta} V^\eta$.

In differential geometry, however, covariant derivatives are objects sending tensor fields to tensor fields (where "tensor field" is a section of a vector bundle of the form $(TM)^{\otimes n}\otimes (T^* M)^{\otimes m}$). How are these two things consistent?
10:55 AM
@glS You can apply $\partial_\mu$ to a map $C^\infty(M)\to\mathbb{R}$ and you can multiply it by a bunch of numbers $\Gamma^\nu_{\mu\eta}$, no?
$\nabla_\mu V^\nu$ is just what $\nabla V$ is in components, there's not really any magic here (and if a differential geometer wants to write down the actual expression for the field in coordinates, they'll use exactly the same expression the physicist uses)
@ACuriousMind well, sure, but it still seems weird to say it acts on a "contravariant tensor field", no? It instead acts on the coordinates of the coordinates of a vector fields. Language aside, so what's the general idea of this? We define $\nabla_\mu V^\nu \equiv \mathrm dx^\nu (\nabla_\mu V)$ with $V=V^\nu \partial_\nu$? I've always been fine with this, but then I get weird stuff applying it to things like the covariant basis itself, as per the question above.
Okay, so $\nabla_\mu$ is the covariant derivative along the $\mu$-th coordinate vector field (i.e. the covariant derivative along $\partial_\mu$ at every point.
The covariant derivative of a vector field $V$ is another vector field, $\nabla_\mu V$ (note this is not coordinate notation, the $\mu$ is not "free")
my understanding is that $\nabla_\mu\equiv\nabla_{\partial_\mu}$ with $\{\partial_\mu\}_\mu$ local basis for the tangent space
and this vector field has components: $\nabla_\mu V = (\nabla_\mu V)^\nu \partial_\nu$, with $(\nabla_\mu V)^\nu = \nabla_\mu V^\nu = (\partial_\mu + \Gamma^\nu_{\mu\rho})V^\rho$
@glS yes, that's what I mean
@ACuriousMind sure, this checks out (modulo small typo on the indices in the last expression I think). What happens though if you apply this to the covariant basis? Say, let $\vec S_\alpha\equiv\partial_\alpha \vec R$ denote the covariant basis, defined from the "position vector" $\vec R$ (I guess this is to be understood as a local parametrisation of the manifold). Then what's $\nabla_\mu \vec S_\alpha$?
11:09 AM
The general idea is, you're just taking a basic (partial) derivative, but now it's a (partial) derivative of a vector field, and in general the basis can depend on position as well as the components so you now need to differentiate the basis vectors as well, but what the heck does the (partial) derivative of a basis vector mean? We don't know, so we just say it's a linear combination of the basis vectors at that point, and the unknown coefficients are called Christoffel symbols.
This is what your (2) is really saying despite the confusing notation, if you re-wrote (1) and the very first term in (2) with a partial derivative it would make more sense, and then defined the coefficient in the last equality in (2) as $\nabla_i Y^j$ it would agree with what you wrote above, but on manifolds they start using this $\nabla$ notation everywhere so it can get confusing
@glS I don't know what a "covariant basis" in this context is, or where $\vec R$ is supposed to come from for a general manifold
@ACuriousMind I understood it as the inverse of a local chart. Which in fairness is a standard way to define a canonical basis for tangent spaces in a differential geometric context as well, I think. Given a chart $(U,\phi)$, $\phi:U\to\mathbb R^n$, we can define a basis for $T_p M$, $p\in U$, as $\partial_i|_p=\phi^* e_i$ with $e_i$ canonical basis on $T_p M$... I think
sure, coordinates define (co)vector bases
but what's $\vec R$?
where did you get this idea from? Why are we using some weird $\vec S_\alpha$ and not just the perfectly fine vectors $\partial_\alpha$?
11:15 AM
not that they say it explicitly in the lectures I was looking at. They take an informal approach and just call "position vector" there. This is a way I though one could make sense of it
if you have a "position vector" you might be looking at an embedded manifold and then you need to carefully distinguish between extrinsic and intrinsic viewpoints
@ACuriousMind mostly, because I was trying to make sense of youtu.be/jQTm-YyKWs0?t=1786. Which look like pretty good lectures on a few topics. Though now I'm starting to have doubts (but remain hopeful)
@ACuriousMind yes, that's an assumption. Sorry, forgot to mention it here
I'm not sure (4) i.e. what he writes on the board there makes any sense
@bolbteppa you mean the double meaning of $\partial_i$ there? yes, I suppose that gets confusing. Maybe it's better to use a different notation for tangent space basis here
@glS ah, there we have it - you need to be careful - the $\cdot$ in that lecture is the Euclidean dot product in the ambient space, not something intrinsic to the manifold
11:26 AM
ok, well, at least I'm not the only one having trouble with it. I feel slightly less dumb. Thing is, the whole framework he uses seems pretty self-consistent. But I think things start to get weirder (from my perspective) when the standard rules of covariant derivatives are applied to co(ntra)variant *vectors*, rather than just their components. I.e. the standard rule $\nabla_\mu X_\nu=\partial_\mu X_\nu - \Gamma^\eta_{~\mu\nu} X_\eta$ makes sense, as mentioned above.

But if I try to apply it to something like $\vec S_\mu$ (covariant basis vector), I'd get zero
@ACuriousMind that last term remains inconsistent btw. Is there even a way to "factor" $V^\rho$ like that?
The covariant derivative has the Leibniz rule on tensors
if you have a scalar field $f$ and a vector field $V$, $\nabla (f V) = (df) V + f (\nabla V)$
@glS I wrote it wrong, it's just $\partial_\mu V^\nu + \Gamma^\nu_{\mu\rho}V^\rho$
The components of a tensor are a scalar field
that is whence the formula comes from
@ACuriousMind sure. But it makes sense also more generally, no? Just defining the contravariant basis via the metric tensor (assuming Riemannian space)
@Slereah which formula are you referring to exactly? I'm fine with Leibniz rule here. But from that you get the standard local expressions $\nabla_\mu\vec S_\nu=\Gamma^{\eta}_{~\mu\nu} \vec S_\eta$ if $\vec S_\mu$ denote the local (covariant) basis for the tangent space. The problem is that this seems to contrast what you get applying the standard "physics rules" of how $\nabla$ acts on objects with raised/lowered indices, because then you'd seemingly get $\nabla_\mu\vec S_\nu=0$
Your vector field can be decomposed in a basis
$$V = V^\mu e_\mu = V^0 e_0 + \ldots$$
Here, each $V^\mu$ is a scalar field, and each $e_\mu$ is a vector field
11:33 AM
@glS okay, you've got me, I'll write an answer :P
So you have, by linearity, that $\nabla V = \sum_{\mu} \nabla( V^\mu e_\mu)$
By the Leibniz rule, this gives you
$$\nabla V = \sum_{\mu} (dV^\mu) e_\mu + V^\mu (\nabla e_\mu))$$
$dV$ is the classic differential of a scalar field, so that it is just some vector in a basis
$dV^\mu = \partial_\nu V^\mu e^\nu$
$\nabla e_\mu$ on the other hand is gonna give you a rank two tensor for every member of the basis
Well, "tensor"
since it's not actually a tensor
@ACuriousMind I feel (unironically) honoured to "have got you" =)
@Slereah I understand this. Question is, what happens if you apply these rules to co(ntra)variant vectors?
sorry, that's not it
Mostly the same thing?
you are applying it to vector fields already. I should have said, how are these rules consistent with the ones often used in physics literature for things like $\nabla_\mu V^\nu$ or $\nabla_\mu V_\nu$, where here $V^\mu$ and $V_\mu$ are components of some vector fields/differential 1-form. If one keeps applying these rules to the local basis (which is seemingly a "covariant object"), you get to zero
Sign will change, 'course
An easy way to see why is to try $\nabla \omega(V)$
Apply the Leibniz rule
11:43 AM
content is the one mentioned in this question btw: physics.stackexchange.com/q/668175/58382
$$\nabla(\omega V) = \nabla(V) \omega + \nabla(\omega) V$$
But also since this is a scalar, $\nabla (\omega V) = d(\omega V)$
If you work out the calculations, it should tell you that the Christoffel symbols are of the opposite sign for dual vectors
@Slereah yes, that's not my point. My point is: if $\vec S_a$ is a local covariant basis, I'd write $\nabla_a \vec S_b=\Gamma^c_{~ ab}\vec S_c$. However, if I apply the rule $\nabla_a V_b = \partial_a V_b - \Gamma^c_{~ ab} V_c$ to $\vec S_a$, I'd get zero, because the Christoffel symbol can be defined, in this context (embedded manifolds etc), as $\vec S^c\cdot \partial_a \vec S_b=\Gamma^c_{ab}$
clearly something's wrong here. My money being on some subtlety associated to applying these rules to vector fields themselves, rather than their components.. but that's what they do in the lectures linked above, and the guy seems to know his stuff, so he might be using a different definition of $\nabla$ or something?
11:58 AM
actually, no, we don't get zero. We get something normal to the surface. It's $\vec S^c\cdot\nabla_a\vec S_b$ which would be zero
If you want to know about the derivatives of vectors in a submanifold you need to look into uuuuh
what is it called
Fundamental forms
O'neill has a big old chapter on it
which is really the main issue I feel. The covariant derivative should give (co)vector fields by definition. This guy, by following seemingly standard rules gets to the conclusion that $\nabla_a \vec S_b=\vec N B_{ab}$ with $\vec N$ normal field. So what he's doing corresponds to some other definition of covariant derivative
Suppose a motorcycle starts at $v_i$ and then skids over a distance of$ d$, ending the skid at $v_f$. Find ${v_f}^2 - {v_i}^2 $ in terms of the distance $d$ and the deceleration a.
Answer assumed that acceleration is constant. I don't see that acceleration is constant is mentioned in the question. Does to skid mean that acceleration is constant?
@glS I needed to make sure, but it's really just what I said at the beginning - you're confusing the intrinsic and extrinsic viewpoints in ways that are simply inadmissible
@ACuriousMind will not admit it
12:10 PM
@ACuriousMind mh.. ok, that makes sense. But are you then saying that what he calls "covariant derivative" is just a completely different object than the covariant derivative in the differential geometric sense? Because this "covariant derivative in the ambient space" gives zero when applied to the covariant basis
@glS It's the covariant derivative in the ambient space
It's not a "completely different object", but it's not the covariant derivative on the surface
you just can't compare the situation of an embedded manifold with an abstract one - the embedded case will have a lot of objects that don't make any sense in the abstract case
but if this was a standard covariant derivative in the ambient Euclidean space $\mathbb R^n$, then wouldn't it amount to the standard directional derivative, being the space flat and all? Then why should I get zero when differentiating covariant basis vectors?
Say, if I think of $\vec S_\theta$ as a (local) tangent vector field on a sphere, its "covariant derivative in the ambient space" should be just it directoinal derivative in $\mathbb R^3$, which doesn't sound like it should be zero?
@glS no, it's not zero, but it's normal to the surface
ah, yes, I keep getting confused on that.
very similar to how in mechanics you'll often find the statement that $\dot{r}$ is orthogonal to $r$
12:23 PM
ok, so his covariant derivative is just the standard directional derivative in the ambient space?
depends on your coordinates! (remember that Christoffels can be non-zero even in flat space)
the covariant derivative even in flat space is a notion distinct from the directional derivative
just look it up mang
there's plenty of introductions to that
@ACuriousMind ah! Yes, that's it I think. The ambient space is always flat, which doesn't however mean $\nabla$ is just the directional derivative. We remove one "source of weirdness" (the non-flatness) but switching to the ambient space, but not the non-orthogonality of the coordinate system
@Slereah the problem here is that they chose to look it up but in two different places, one using extrinsic language and one intrinsic language :P
That's all part of growing up
You have to be able to piece together two different formalisms
a functor mb
12:30 PM
well, the problem is that they're really two different formalisms (like using $\nabla$ for two different notions)
I can understand the confusion
@ACuriousMind I guess I was also tricked by the fact that the answers are the same as long as we deal with things like $\nabla_\mu V^\nu$ with $V^\mu$ components of a vector field. But I can see now that those were used when he was discussing the case of a flat embedded surface. I suppose then those rules are then only true when projecting back on the tangent space in the non-flat case (otherwise we'd get terms in the normal direction)
@glS careful, "flat" is also ambiguous in the embedded case ;) You probably mean a surface with vanishing second fundamental form
you can have extrinsically flat manifolds with non-zero intrinsic curvatures and vice versa
(though not necessarily when embedded in Euclidean space)
cf. the cylinder
(what we see here is that if you thought diff. geo. might be easier when considering manifolds as embedded in Euclidean space, it isn't, really :P)
@ACuriousMind ah, ffs. What does "extrinsically flat with non-zero intrinsic curvature" mean? Like, an Euclidean space with a non-orthogonal choice of coordinate system?
12:48 PM
@glS you can't have that in Euclidean space, but if you have an intrinsically curved space, then you can embed other intrinsically curved manifolds into it but with vanishing extrinsic curvature (i.e. vanishing second fundamental form/shape operator)
the simplest example I know is $S^2$ into $S^3$ - both are intrinsically curved, but the embedding has no extrinsic curvature if you choose a "great sphere" (analog to a great circle on $S^2$)
the other way around is the cylinder Slereah already mentioned - its usual embedding in $\mathbb{R}^3$ is extrinsically curved, but it's intrinsically flat.
mh.. reading https://math.stackexchange.com/a/1711963/173147, I can understand "extrinsic curvature" as a topological property. Locally cannot be detected, but you can find weird stuff going around some loops (or you just have loops that were impossible if the space was really flat, etc). So I can understand the cylinder from this point of view. But this is "extrinsically nonflat, intrinsically flat" I guess.

Your example means that if I embed $S^2$ into $S^3$, there is no extrinsic curvature. What does it mean precisely? Does it refer to the fact that topologically this embedding of $S^2$
so in the embedding $S^2\to S^3$ any loop is contractible, fundamental group is trivial, but still you have an intrinsic curvature
(I mean, it still sounds weird, but I suppose some weird embedding in $S^3$ might make it possible...)
@ACuriousMind does this embedding have a name?
1:07 PM
sorry to bring up a tangential question, but is there a good reason why someone would do an embedding of $S^2$ into $S^3$ or some arbitrary manifold? I had always assumed that embeddings were always into $\mathbb{R}^N$ just because it's easy
Some of them suggest themselves
if you have $A \times B$, there's an embedded $A$ manifold
also plenty of submanifolds have nice properties in general
@glS this specific embedding not, but in general such extrinsically flat embeddings are called totally geodesic submanifolds
(because the geodesics on the submanifold agree with the geodesics of the ambient manifold)
this points to another way to think about extrinsic curvature/second fundamental form: it measures how much the notion of distance/geodesic differs between the ambient metric/LC connection and the intrinsic metric/LC connection
@DanielUnderwood usually you don't get embeddings by starting with $X$ and looking for an embedding into $Y$, but you start with a submanifold $X\subset Y$ and when you consider $X$ as an isolated manifold you have the natural embedding given by the inclusion
1:33 PM
bonjour a todos
2:31 PM
A: Are differential geometric and physics conventions for covariant derivatives consistent?

Vincent ThackerI have read the same book as the one you linked, and I can say that it does have some technical inaccuracies that slowly build up to the confusion that you are having right now. I will try to list them in order. I will denote the standard covariant basis by $\mathbf{e}_i$, and I will use Greek le...

@ACuriousMind they make a good point in this answer. The covariant vectors $\mathbf e_j$ are still vector fields, so shouldn't we apply the formula with the plus sign (i.e. the one for "contravariant objects") for the covariant derivative?
3:18 PM
@ACuriousMind where in physics are we not going to be able to embed things into a higher Euclidean/Lorentzian manifold (Whitney etc) and just do normal calculus
Aren't the counterexamples super exotic and terrible
3:36 PM
How do i use the t test between two numbers that has different df?
Even in something like local gauge invariance the 'comparator' really just turns it back into basic calculus
3:54 PM
@bolbteppa so, just to clarify, you disagree with Vincent's answer I take it. So what are you saying $\nabla_i \mathbf e_j$ should be here?
It's not common to do it like that but if you want to formally extract the $\nabla$ from $\partial_i \mathbf{A} = (\nabla_i A^m) \mathbf{e}_m$ and pretend it's some abstract operator which satisfies the product rule and acts one way on the coordinates $A^i$ and another way on the fields $\mathbf{e}_m$ you can do it but it's trivial in context
This is basically a simple calculus thing that's made to look really difficult with multiple spaces and isomorphisms and you end up defining new things that are really just trivial
@bolbteppa so... you are saying $\mathbf e^k\cdot \nabla_i\mathbf e_j=0$?
because for me, the way $\nabla$ is defined in the abstract, and thus how it acts on vector fields, seem perfectly natural and consistent. But then the covariant basis vectors $\mathbf e_i$ are again vector fields, and so it makes sense that $\nabla$ acts on them in the usual way (i.e. with the plus sign, essentially)
I would not recommend doing it this way, even calling $\partial_i$ a basis I would not recommend...
@bolbteppa ok, but that's not answer. So you are saying $\mathbf e^k\cdot\nabla_i\mathbf e_j=0$ is technically correct, but not recommended?
4:10 PM
If you defined $\nabla_i \mathbf{e}_j = 0$ then why wouldn't that be zero too
@bolbteppa well, I guess the question is whether $\nabla_i\mathbf e_j=0$ is true for an arbitrary connection and covariant basis. Or if you want to start from that expression, whether the operator $\nabla_i$ thus defined is indeed a valid connection
I mean, if $\nabla_i\mathbf e_j=0$, doesn't that mean that you are dealing with a trivial connection?
In mathematics and physics, the Christoffel symbols are an array of numbers describing a metric connection. The metric connection is a specialization of the affine connection to surfaces or other manifolds endowed with a metric, allowing distances to be measured on that surface. In differential geometry, an affine connection can be defined without reference to a metric, and many additional concepts follow: parallel transport, covariant derivatives, geodesics, etc. also do not require the concept of a metric. However, when a metric is available, these concepts can be directly tied to the "shape...
Basically you don't use Christoffel symbols if it's not true
I don't understand what you mean. If it's true, the Christoffel symbols vanish. Is that what you mean with "don't use"?
"The Christoffel symbols are most typically defined in a coordinate basis, which is the convention followed here. In other words, the name Christoffel symbols is reserved only for coordinate (i.e., holonomic) frames. However, the connection coefficients can also be defined in an arbitrary (i.e., nonholonomic) basis of tangent vectors $\mathbf{u}_i$ by ..."
4:39 PM
A: What is the meaning of non-coordinate basis?

Bence RacskóIntegral curves of non-coordinate (anholonomic) basis vectors also exist, they just don't form a coordinate system. This might be a bit difficult to swallow, but the heart of the issue in a coordinate system, the coordinates are independent. Here's a direct example: Consider polar coordinates ...

That's a good example
Q: Proper formatting of links

ACBDoes it make any difference in using https://physics.meta.stackexchange.com/q/13855/305718 instead of https://physics.meta.stackexchange.com/questions/13855/proper-formatting-of-links ? (Link is of this post as an example)

5:33 PM
To see this in an easy fashion, using the position vector $\mathbf{r}(x^i)$ for curvilinear coordinates $x^i$, the $\mathbf{e}_i = \frac{\partial \mathbf{r}}{\partial x^i}$ form a basis, and so if you define $\mathbf{A} = A^i \mathbf{e}_i$ then
$$\partial_j \mathbf{A} = (\partial_j A^i) \mathbf{e}_i + A^i (\partial_j \mathbf{e}_i) = (\partial_j A^i) + A^i (\Gamma_{ij}^k \mathbf{e}_k) = (\partial_j A^i + \Gamma_{kj}^i A^k) \mathbf{e}_i$$
suggests we *define* $\nabla_j A^i = \partial_j A^i + \Gamma_{kj}^i A^k$. From that $\partial_j \mathbf{e}_i = \Gamma_{ij}^k \mathbf{e}_k$ we can find the
(Some hats/tildes missing on a few at the end)
@bolbteppa this is an interesting example and way to see how to get a non-symmetric $\Gamma$, but I'm not seeing the connection (ha!) with the issue at hand. The problem is what $\nabla_i\mathbf e_j$, the action of the covariant derivative on a vector field with a lowered index, should be taken to mean
following that standard derivation, you'd get the standard answer, that is, $\nabla_i \mathbf e_j=\Gamma^k_{ij}\mathbf e_k$
I already explained, we have $\partial_i \mathbf{A} = (\nabla_i A^m) \mathbf{e}_m$ above in both symmmetric and un-symmetric cases right? In both cases we can formally just pull off $\nabla_i$ and say $\partial_i \mathbf{A} = (\nabla_i A^m) \mathbf{e}_m = \nabla_i (\nabla_i A^m \mathbf{e}_m)$ as long as $\nabla_i \mathbf{e}_m = 0$
Clearly it's just a trivial notation choice
yes, but also, $\mathbf e_i$ is an example of such a $\mathbf A$, so applying the same formula you'd get $\nabla_i\mathbf e_j\neq0$
idk, I'm probably being dumb here. I'll think about it later with a fresh mind
5:50 PM
Since it's just a notational choice, we could fix it with more notation, but you're right that shows how absurd it is to use this notation
Notice I just used basic partial derivatives and vector norms and managed to find the scary case of non-symmetric connections in non-holonomic bases in two seconds
Technically $\mathbf{A} = \mathbf{e}_i$ is a vector field $\mathbf{A} = A^i \mathbf{e}_i$ which satisfies $\partial_j A^i = 0$ because $A^i = 1$ happens to be the case for all $A^i(x^1,x^2,...)$
6:14 PM
A confession: I’ve never been able to get a good grasp on wtf a connection is
Whenever I’ve tried to look it up on Wikipedia my brain just throws a “does-not-compute” error
A connection tells you what a vector looks like if you move it, roughly
It's a way to compare the tangent space at two different points
Yeah, something something parallel transport
I’m just never clear on what kind of object it is mathematically.
nCatLab says it’s a functor, which probably helps explain why I find it hard to grok
I mean it's nlab
Everything is a functor over there
The connection tells you where a vector ought to be pointing if it's on a geodesic, roughly?
@Slereah Lol, you have a point
Applying the addition functor to the 1 category
mapped into the 2 category
6:23 PM
Hmm. So you tell a story for how tangent vectors transform if you move along geodesics.
And therefore how it’ll transform under arbitrary paths?
When you take a partial derivative of a vector field,
$$\partial_i \mathbf{A} = \partial_i (A^j \mathbf{e}_j) = (\partial_i A^j) \mathbf{e}_j + A^j (\partial_i\mathbf{e}_j)$$
you have to ask, what does $ (\partial_i\mathbf{e}_j)$ mean? Whatever it means, we can write the result as a linear combination of the basis vectors at that point, so we just set
$$ (\partial_i\mathbf{e}_j) = \Gamma_{ji}^k \mathbf{e}_k$$
that's pretty much all there is to it. That's how they calculate the Christoffels in this example for example:
In mathematics and physics, the Christoffel symbols are an array of numbers describing a metric connection. The metric connection is a specialization of the affine connection to surfaces or other manifolds endowed with a metric, allowing distances to be measured on that surface. In differential geometry, an affine connection can be defined without reference to a metric, and many additional concepts follow: parallel transport, covariant derivatives, geodesics, etc. also do not require the concept of a metric. However, when a metric is available, these concepts can be directly tied to the "shape...
You may remember the holonomic definition of the connection
Not really :P
we agree that $\partial_i \mathbf e_j=\Gamma^k_{ij}\mathbf e_k$, right? Then, you say, define the action on $\nabla$ of components of a vector field as $\partial_i (A^j\mathbf e_j)=(\nabla_i A^j)\mathbf e_j$. But then, being $\mathbf e_i$ a vector field in and of itself, we'd have $\Gamma^k_{ij}\mathbf e_k = \partial_i \mathbf e_j = (\nabla_i 1)\mathbf e_j$ (because the coefficients of the decomposition of $\mathbf e_i$ on itself are the delta). So we must always have $\Gamma=0$?

I mean, I'm fine with this implying $\Gamma=0$, but that is not always the case. Even just in polar coordinates
Tho invoking holonomy starts to shed some light
6:27 PM
I'm trying to explain but tbh the definition of a connection is something I fully understand when I find that one good paper about it
I just read it once in a while and remember what a connection is
Lol, fair
@bolbteppa I don't mind the notation being "hard" or "not recommended" here. My point whether it makes sense at all/it's consistent. Which doesn't seem to be the case
We can then use that definition re-write the partial derivative of $\mathbf{A}$ as $\partial_i \mathbf{A} = (\nabla_i A^j) \mathbf{e}_j$ where I defined $\nabla_i A^j$ above.
The little horizontal bundle to the bundle
I’ve never been interested in GR stuff but I still would like to understand connections in QM
6:27 PM
We can then say a vector is parallel transported if $\nabla_i A^j = 0$, i.e. the partial derivative of the vector field including it's basis does not change, $\partial_i \mathbf{A} = 0$. It's trivial to see parallel transport means the vector doesn't change, although it's components can change the changes have to be countered by changes in the basis to ensure $\nabla_i A^j = \partial_i A^j + \Gamma_{ki}^j A^k = 0$ holds
Berry phase stuff
this is all I know of berries
When you abstract to manifolds, where we pretend we can't even define a derivative because we have to evaluate $\mathbf{e}_j(x)$ at $x$ and at $x+dx$, i.e. vectors in two different tangent spaces, we have to come up with another definition to say we moved the vector in the tangent space at $x+dx$ to the one at $x$ without changing it, what's the obvious definition
We parallel transported it, and it just so happens to agree with the definition of parallel transport I gave above
Instead of translation a point you flow it with a lil diffeomorphism
or equivalently move along a curve
In the QM context I mostly know stuff like: take a 2-level system. It’s (pure) state space is the Bloch sphere
6:29 PM
curve deriving from a vector field
etc etc
@glS It's not $1$ it's a vector field $\mathbf{A}$ defined at each point $(x,y,z)$ such that it's component fields $A^i(x,y,z)$ happen to be equal to $1$ so that $\nabla_i A^j(x,y,z) = \partial_i A^j + \Gamma_{ki}^j A^k = 0 + \Gamma_{ji}^j \delta^{ji} $
That picture specifies each state up to phase. To incorporate that info, we could assign a unit tangent vector to the sphere at a particular point, and interpret that as the phase
And then the weirdness is that if you smoothly change the state—say, rotate the Bloch around the y-axis
Then the state you get upon making a full rotation, will still be the state you started with but can have a different phase
@bolbteppa mmh I mean the coefficient along the direction $\mathbf e_i$ of $\mathbf e_i$ is not a vector field, is it. If anything, you are saying $\mathbf e_i(x)=f^i(x)\mathbf e_i(x)$ where $f^j(x)=\delta_{ij}$ for all $x$ (or at least locally in some chart I suppose). But still, $\nabla_i f^j=0$ I'd expect?
Last term, if $\mathbf{A} = \mathbf{e}_j$, should be $\nabla_i A^j(x,y,z) = \partial_i A^j + \Gamma_{ki}^j A^k = 0 + \Gamma_{ki}^j \delta^{kj} = \Gamma_{ji}^j$ right
So the unit tangent vector doesn’t have to return to its initial direction
This paper is where I saw it described like that. Need to read it more carefully: aapt.scitation.org/doi/10.1119/1.16809
6:40 PM
That is right except $\nabla_i f^j \neq 0$ if $\mathbf{e}_i(x)$ has non-zero derivatives. If
$$\mathbf{e}_i(x) = \delta_{ij} \mathbf{e}_j(x) = f^j(x) \mathbf{e}_j(x)$$
then $f^j = \delta_{ij}$ is the coefficient and so
$$\nabla_k f^j = \partial_k f^j + \Gamma^j_{lk} f^l = \partial_k(\delta_{ij}) + \Gamma_{lk}^j \delta_{il} = 0 + \Gamma_{ik}^j $$
Since $i$ is really a fixed value like $i = 3$ or something it's fine from the perspective of index notation
@gls there’s a passing comment in that Choice = Signal paper that seems interesting
“ We note in passing that making a non-selective measurement in basis b is equivalent to performing a random unitary transformation which is diagonal in the b basis.”
So presumably part of the story here is that, while infinitely many such unitaries are possible
The unitaries which are diagonal in one of the pre-selected bases are a subset of arbitrary unitaries
So if Alice knows that, then it’s not so implausible that such a unitary has a measurable effect
6:56 PM
ugh. Ok, yes, I did it 3 times messing up the indices in 3 different ways, but I'm finally there. So this seems consistent, fine, we get what we should. What I'm still not clear about is: is this still a connection in the usual formal sense?
As far as I understand, a Levi-Civita connection is *defined* by its action on coordinate basis vector fields (https://en.wikipedia.org/wiki/Levi-Civita_connection#Christoffel_symbols), So if we now say $\nabla_i\mathbf e_j=0$, we would be saying $\nabla_i=0$. So would this be a non Levi-Civita connection or something?
I don't see where $\nabla_i = 0$ comes from, I don't think this notation is a good idea at all if it's causing this much trouble, do you agree with everything else apart from $\nabla_i \mathbf{e}_j$
@Semiclassical mmh. Performing a random unitary on what exactly?
If you're working with connection in a serious manner, I recommend that you keep using it consistently as a function
I find it useful
I mean, I'm just trying to understand if all this makes sense upon some weird choice of connection or something. I don't care about the practicality of using it rn, I just want to understand what I don't. I'm saying $\nabla_i=0$ because a Levi-Civita connection is characterised by its action on a coordinate basis, i.e. $\nabla_i\mathbf e_i=\Gamma^k_{ij}\mathbf e_k$ with some correspodning Christoffel symbols.
In our context, maybe these are not the same Christoffel symbols or something, that's fine. But still, a LC connection should be entirely defined by the way it acts on $\mathbf e_i$. S
for example, if it turns out this notation corresponds to a choice of non-levi-civita or other weird connection, that would be interesting as an example of such a thing to keep in mind
It's really supposed to be trivial, if it's not trivial forget about it as bad notation, but I have no idea where $\nabla_i = 0$ is coming from
7:10 PM
I'm not even sure what it's supposed to mean
it probably means I have to sleep lol
@glS the second subsystem
My schema right now: Alice does a measurement on the combined system to prepare it, Bob picks a unitary to act on the second subsystem, and Alice makes another measurement on the combined system to get info on what unitary Bob picked
In general, I can’t imagine that you could do much with an arbitrary unitary acting in between. But if the unitary is constrained (must be diagonal in one of the preassigned bases) then that’s more plausible
7:49 PM
Here's a simple concrete example: Consider $\mathbf{A} = \hat{i}$. Clearly it's derivatives should vanish. We can put this in polar coordinates by taking the position vector $\mathbf{r}(x,y) = x \hat{i} + y \hat{j}$, putting it in polar coordinates $\mathbf{r}(r,\theta) = r \cos \theta \hat{i} + r \sin \theta \hat{j}$, a basis is given by
\mathbf{e}_r &= \frac{\partial \mathbf{r}}{\partial r} = \cos \theta \hat{i} + \sin \theta \hat{j} \\
\mathbf{e}_{\theta} &= \frac{\partial \mathbf{r}}{\partial \theta} = - r \sin \theta \hat{i} + r \cos \theta \hat{j}.
@Semiclassical Where's the difficulty with the connection in that example compared to the above explanation

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