The_Vman:
@gentle osprey yes
#point-set-topology
knotty pasture
gentle ospreyBOT
The_Vman:
gentle ospreyBOT
The_Vman:
knotty pasture
ok but $U\cap U_i$ is open
gentle ospreyBOT
Davide:
unkempt bolt
Yes
knotty pasture
then $\bigcup_i U\cap U_i=U $ is open
gentle ospreyBOT
Davide:
unkempt bolt
Exactly
unkempt bolt
No problem
feral dragon
Do you guys think I can complete Munkres Topology in a week ? lol well if not EVERYTHING, certainly most of it
gritty widget
no
coarse kestrel
I don't think so, but I certain wish I could complete it in a week
nimble pebble
theres probably someone out there who can
but theres someone out there who went to college at age 12 so take that with a grain of salt
gritty widget
i mean
point set topology is dry enough that i can see why someone would want to do it in a week lmao
gritty widget
Yeah I do it every wendesday
uncut surge
Let $M$ be a smooth manifold of dimension $n$, and $\mathcal{U}$ any open cover of $M$, in particular $\mathcal{U}$ is not necessarily good. Consider then the geometric realization of the nerve of $\mathcal{U}$. Is this homotopic/weakly equivalent to a space of covering dimension $\leq n$?
gentle ospreyBOT
Lartomato:
uncut surge
So in general, the nerve theorem tells me that if my open cover is a good one, doing this nerve->geo.realization thing automatically gives me a space which is weakly equivalent to my original space. This doesn't need to happen if my open cover is badly chosen, but what I'm hoping for is that even an arbitrarily bad cover will still give me something which is "essentially" at most n-dimensional
lots of complicated spoopy words, maybe better on math.stackexchange, idk
uncut surge
https://math.stackexchange.com/questions/3800067/can-the-geometric-realization-increase-the-dimension#3800077 wow it's wrong
i'm done thanks
marsh forge
wait
i cant tell what the answer is saying
it says no it cant but then provides an example which makes me think its trying to show that it can
okay yes its showing that it can increase dimension
they wrote their answer weirdly
i think i would have guessed this because it feels like homotopy memes like geometric realization dont play well with homotopy variants
uncut surge
yeah it's quite a bummer
i'm just trying to fix the proof in a paper i'm reading
which is like "for the statement of this proof, we just use corollary 5.4 in this other paper : - )" but swiftly ignore that they want to do something for all open covers of a manifold, but the corollary only works for good open covers
now idk what to do but at least this makes sense to me now
and even if the proof is a bit spoopy, the example he gives is convincing
marsh forge
Are you sure the author isnt using only good covers
uncut surge
they're proving that a thing is a homotopy (co-)sheaf, so they should be proving that the homotopy (co-)sheaf conditions hold for arbitrary covers
the best thing i could hope for is some kind of refinement argument, along the lines of "if the sheaf conditions holds for all good covers, then it also holds for non-good covers"? i don't know if that's likely, though
it's just a preprint in any case, and i already sent the author an e-mail, so i'll just hope they respond to me π€·ββοΈ
wicked trout
Iβm probably just missing something simple, but could someone point me in the direction of proving that S^2 x S^1 is a prime 3-manifold (ie canβt be expressed as a non-trivial connected sum)?
uncut surge
Hm, I'm not at all familiar with prime manifolds, but I scrolled through the definitions and results a bit, and it seems that one line of argument could be: If $S^2 \times S^1$ was not prime, it would be equal to a connected sum $M # N$ with neither $M$ nor $N$ homeomorphic to $S^3$. However, consider the fundamental group: $\mathbb{Z} = \pi_1(S^1 \times S^2) = \pi_1(M # N) = \pi_1(M) * \pi_1(N)$, where the last star denotes the free product. But $\mathbb{Z}$ is not the free product of two nontrivial groups, so one of $M$ and $N$ must be simply connected
gentle ospreyBOT
Lartomato:
uncut surge
And now a simply connected closed 3-manifold is $S^3$ by hitting it with the PoincarΓ© conjecture 
There's probably a way to see this without resorting to a Millenium problem, but at least this is somewhat easy to remember
gentle ospreyBOT
Lartomato:
wicked trout
Looks good to me, thanks @uncut surge ! I forgot about that minor conjecture some dude proved recently 
uncut geyser
@gritty widget not dry 
fathom ruin
What is a good proof for this theorem?
Through a point outside a line, there is exactly one line parallel to the given line
chrome dew
lmfao
fathom ruin
Nvm I thought of one
chrome dew
sure you did
dim meadow
Lmao
shut marsh
"parametrized by proper time" is just physics-speak for "parametrized by arc length" right?
ugh, why am I doing this to myself
Well, from my experience, the fraction of mathematicians familiar with physics terms is much larger than that of physicist familiar with mathematical terms...
gritty widget
what parts of analysis in particular are helpful for learning topology?
should I ask soft questions in another channel btw
Yea
marsh forge
yeah like first few chs of rudin at most
gritty widget
Alright nice, Iβve just started reading rudin
cursive flume
how can one define the levi civita tensor coordinate free?
unkempt bolt
I can't answer this, and really have no background in diff geometry, but there's this https://en.wikipedia.org/wiki/Levi-Civita_connection#Fundamental_theorem_of_(pseudo)_Riemannien_Geometry
Where it says "so we find the Koszul formula" it gives an expression that only uses the metric
cursive flume
this is levi civita connection,i'm looking for the tensor
unkempt bolt
Rip my ability to read
shut marsh
@cursive flume I have been reading "Differential geometry, connections, curvature and characteristic classes" recently.
It's explained very nicely there
cursive flume
can you give a screenshot please?
shut marsh
@cursive flume https://m.atx.name/difftu.pdf
See Definition 6.4 onwards
shut marsh
Also, this ncatlab quote
As every concrete component expression, Christoffel symbols may be useful in certain computations. Unfortunately, almost every textbook on gravity in theoretical physics follows the long-outdated tradition of describing (or not describing) the entire notion of connections on tangent bundles without introducing these conceptually but just describing the Yoga of how to handle the Christoffel symbol component. This way their main effect to science nowadays is to make it harder for students of theoretical physics to understand what is really going on in the universe. Itβs all so simple. Speaking always in terms of Christoffel symbols and never in terms of the abstract notion of connection makes it all so hard.
Is the wiki picture (https://en.wikipedia.org/wiki/Torsion_tensor) a good way of thinking about torsion?
Tu says "there does not seem to be a good reason for calling T(X,Y) the torsion", but the picture looks so suggestive
chrome dew
it's a relative tensor @cursive flume and you can fix that by cancelling out the difference with another relative tensor like the sqrt of the determinant of the metric tensor
cursive flume
yes but i've only seen it being done in charts
i can do it in charts aswell
but i'd be interested in global def.
chrome dew
not sure what you mean by that
cursive flume
i'd be interested in a definition as a tensor field
where if i plug in a basis,i get the components
like a multilinear map on gamma(tm)
shut marsh
@chrome dew Some people have this annoying tendency to write https://en.wikipedia.org/wiki/Christoffel_symbols#General_definition and be like "here, have this, go practice index gymnastics"
cursive flume
for example @chrome dew this is the coordinate independent formulation of riemann tensor:
and if i plug in a basis on each field, i get the components
i'd like something like that for the levi civita tensor
but i just find this:
like this is already plugged in
shut marsh
(note that this is a proof of existence & uniqueness, Tu actually defines the levi-civita connection axiomatically)
chrome dew
you know how to write the determinant coordinate free?
probably some way to do it with the metric tensor that way, I don't know, I have no problem with the index gymnastics way personally so long as I can show the properties lol
shut marsh
@chrome dew Isn't the usual way to write the determinant without coordinates by saying "let det be a non-zero n-linear alternating map on an n-dimensional vector space"?
That is, an n-form
Of course this defines an entire family of determinants, but I do not think that matters too much
@cursive flume oh lol, I just noticed that you actually want the levi-civita symbols. And you even said so after someone already pointed you to the connection. And I still haven't read properly.
I should go to sleep... :/
shut marsh
@cursive flume Hm, in that case I would try something like "let π be a differential n-form on M such that π(p) is an orthonormal basis of Ξ^n(T_p*M) at every point p of M"
cursive flume
yes but that's locally defined
i'm interested in saying this is a tensor field on gamma tm defined as ...
if you plug in basis on each field, you get the components
like I posted above in the def of riemann
shut marsh
Is it (excuse my potential noobishness)? Like, I just want <π|π> = 1 everywhere.
cursive flume
idk mathematically rigorously,im a physicist
im studying based on a matphys course
and ive been taught all tensor fields are just multilinear maps on the c infty m module gamma tm
shut marsh
lol it can get worse, I am an electrical engineer (well, hopefully, soon)
cursive flume
and if u want components,u plug in a basis
burnt spruce
how can one define the levi civita tensor coordinate free?
@cursive flume you can get covariant derivatives by assuming your manifold gets its metric as a submanifold of euclidean space, differentiating the appropriate object in R^n and projecting onto the tangent space. This is hinted at on wiki "even though the original motivation relied on a specific embedding" ... one place I know that this is explained in in Thorpe's undergrad diff geo book, specifically this chapter : https://link.springer.com/chapter/10.1007/978-1-4612-6153-7_8
I'm not sure how general this is -- I've only seen 'appropriate object' = vector field, but maybe also works more generally (and proof is possibly immediate from formal properties if true). Not a diff geometer.
burnt spruce
I think this is also proven in Do Carmo's Riemannian geometry in the chapter about induced metrics.
shut marsh
Say I have a vector field which vanishes at a point p. I want to find a chart containing p such that the vector field looks like a system of linear ODEs in that chart.
Can I always do that?
This is apparently a thing in geometric control theory, but I can't find any reasonable reference...
gritty widget
what is an open set 
ivory dragon
what level of generality are we talking here?
in metric spaces, an open set is a set where each point has a neighbourhood around it
i.e. it doesnt contain any of its boundary points
unkempt bolt
@shut marsh I have like no experience in ODEs, and even less about them using charts, but would something like $f'(x(t), y(t)) = x(t) * y(t)$ be counterexample at (x, y) = (0, 0) where the derivative vanishes only at the x and y axis in any neighbourhood of zero
gentle ospreyBOT
The_Vman:
ivory dragon
in a more general setting (ie arbitrary topological spaces, not necessarily with a metric), an open set is just a part of the topology you define on the set
at which point it's like asking "what is a vector"; it's an element of the topology, just as a vector is an element of a vector space
(specifically, open sets are subsets of a set such that finite intersections and at-most-countable unions produce more open sets)
(and the empty set and the set itself are open)
(alternatively, open sets are the complements of closed sets)
to get more "concrete", on R, the open sets are at-most-countable unions of open intervals
for exmaple, (0, 3) is an open set
as is (-10, 15) U (25, infinity)
@gritty widget
gritty widget
@ivory dragon topology
ivory dragon
that clears absolutely nothing up.
ivory dragon
are you dealing with arbitrary topological spaces? metric spaces? euclidean space?
ivory dragon
then as i said
to clarify this explanation a bit:
open sets come from the definition of a topological space
topological spaces are built by taking a set (call it $S$) and defining a topology on it (call it $T$)
gentle ospreyBOT
Namington:
ivory dragon
topologies are essentially a collection of subsets of $S$ called ``open sets"
gentle ospreyBOT
Namington:
ivory dragon
the topology satisfies these axioms
that's what an open set is; it's part of how the topology is defined
this is a very "non-concrete" notion, but this is what an "open set" is in full generality
gritty widget
So if it's a subset of the topology it's open, as long as these axioms are met of course.
ivory dragon
yes (ignoring set-theoretic issues)
gritty widget
$S$ is a set $T$ is a topology of $S$ $x \in T$ is open if the axioms are met
ivory dragon
but its worth noting that a given set may have multiple different topologies defined on it
gentle ospreyBOT
Sasuke Uchiha:
ivory dragon
so what counts as an "open set" depends on what your topology is
the "simplest" topology we can put on a set, in a sense, is the "indiscrete" or "trivial" topology
this is the topology where the only open sets are the empty set and the set itself
alternatively, we could go to the other extreme; the "discrete" or "power set" topology
gritty widget
Is P(X) a topology of X?
ivory dragon
sorry
somehow i mixed up indiscrete vs discrete
lmao
i dont think ive ever done that before
anyway, fixed
anyway, these are the two "extremes" of a topology
but thats not the only way we can define a topology
the open sets of the standard (euclidean) topology on R are defined by:
- R and the empty set are open (as always)
- open intervals are open sets
- unions and finite intersections of open sets are open (as always)
by "open interval" here i mean an interval of the form (a, b) where a, b are real numbers
of course, to define this notion we need to order R
hence we often define a metric space
note that i said that (a, b) are real numbers, but (a, infty) and (-infty, b) would also be open sets
because unions of open sets are open
so we can take an infinite union
say, of (1, 3) U (2, 4) U (3, 5) U (4, 6) U (5, 7) U ...
this infinite union gets us (1, infinity)
which is open, since it's just a union of open sets!
in practice, when people think of "open sets", this sort of construction is the intuition they're usually using
but the topological space definition is far more abstract
note, however, that we don't allow taking the intersection of infinitely many open sets
otherwise i couild do something like this:
$(1, 2.1) \cap (1, 2.01) \cap (1, 2.001) \cap (1, 2.0001) \cap \dots$
gentle ospreyBOT
Namington:
ivory dragon
in the limit this is (1, 2]
we can't call this set open
since we can only construct it by intersecting infinitely many sets, which doesnt necessarily produce an open set
and this should match our intuition; i wouldnt call (1, 2] "open", at least on the right hand side
[though note that it's not closed either!]
hopefully that helps you contextualize this
as an aside: as i mentioned, (1, 2] is neither open nor closed
this is actually very common
many subsets in many topological spaces are neither open nor closed
because sets in topologies are just created using some tottally arbitrary union/intersection of subset rules
and there's no reason to expect every subset to be constructible using that
this makes it quite interesting when every subset either is open or closed (or both)
such a topology is called a "door topology", and they're quite uncommon unless you're specifically looking for them
marsh forge
Yeah makes sense
ivory dragon
not much reason to study them
but they make for good exercises in a general topology text IMO
little hemlock
kind of a dumb question, but I want to show that given some set $X$ and $S \subset \mathcal P(X)$, that the topology generated by $S$, $\tau_S$, is the same as the topology generated by the basis $B$, $\tau_B$ given by the set of all finite intersections of elements of $S$. Since $\tau_S$ is the intersection of all topologies on $X$ containing $S$, we have that $\tau_S \subseteq \tau_B$. I can't think of a "nice" way to frame the other inclusion though...\ \
This is what I have in mind: given $U \in \tau_B$, for each $x \in U$ there is $b \in B$ such that $x\in b \subseteq U$. Equivalently, you could say that $U$ is the union of finite intersections of elements of $S$. Then since $S \subset \tau_S$, the finite intersections and unions of those finite intersections of elements of $S$ must be in $\tau_S$. So $U \in \tau_S$.
gentle ospreyBOT
kxrider:
little hemlock
Is there any nicer way to show this? i.e. some way that doesn't make cumbersome, explicit references to the fact that elements of tau_S are the "unions of finite intersections of elements of S?"
marsh forge
Show that the given topology is coarser and finer than the the other
In particular, the generated topology contains all finite intersections
Basicaly there are simple reasons why this must he true and you just write them down
uncut geyser
@ivory dragon why "at most countable unions"?
ivory dragon
actually hold on
that should just be "infinite"
my apologies
i thought it made no difference but i just thought of a counterexample lmao
thanks for the correction
uncut geyser
cocountable makes some diff o,o
ah np
Im p sure you know that tho
was just asking to know what led you to write
ivory dragon
honestly im not sure
sometimes i get caught up in explanations and just dont think
this is why i usually lecture with notes
uncut geyser
lecturing 
nimble cipher
Hi! How do I prove that the topology with a subbase of union of all topologies $\mathcal{T}\alpha$ is the smallest topology containing all $\mathcal{T}\alpha$. Thank you!
gentle ospreyBOT
emphatic_wax:
unkempt bolt
Ok so here $\mathcal{T}\alpha$ is a family of topologies on some space $X$? If so my suggestion is to show that any topology containing all of the $\mathcal{T}\alpha$ is necessarily finer than the one generated by your subbase
gentle ospreyBOT
The_Vman:
nimble cipher
So I will have to show that any other topology containing the family of topologies is finer than the topology generated by the union. Is that what you're saying?
unkempt bolt
Yeah
nimble cipher
That's great. Thank you!
loud scarab
A question on the box & product topologies, my question is about why for the product topology, the basis element can be written as the product of open sets for a finite number of spaces, while the rest are the entire space.
So I understand that to construct a basis element from a subbasis, we can only have finite intersections of subbasis elements to form a basis element, and here the finite set of indices is represented by beta 1, beta 2, ..., beta n , and I understand that there is no restriction on alpha if it is not one of these indices, but what gives us the right to say that U_{alpha} is the entire space if alpha is not equal to one of these indices?
So, specifically, my question is about this part
loud scarab
but why is it the entire space?
marsh forge
Im pretty sure they were just like
Telling you about this result
And in this theorem they formally prove it
Do they define the box topology before your screenshot
Then yeah thats whats going on
Its a common thing in math writing
To casually explain and idea
And then formally state and prove it
Yeah what they are doing
Is describing the basis elements
Bc the comparison is obvious
Once you know thag
Yeah box topology is generated by products of open sets only finitely many of which are nontrivial
Although this is misleading bc this is what we mean by product
loud scarab
@gritty widget so I understand that all but finitely many have to be intersected to generate a basis element but
marsh forge
Did I
loud scarab
so all but finitely many can be different from the whole space
@gritty widget
marsh forge
I might have
Yes i did
Youre right
That makes much more sense
(Me messing this up is a good example of why no one cares about the box topology)
loud scarab
Yes
so from trying to go from this
to this
I use the definition
of preimage
and expand every pi^-1
and I get the last equation yes
but there's still something
that the first equation is a finite product
while the last is infinite
but how could the first be infinite if it is a finite intersection of subbasis elements?
ohhhhh
finally, thats the piece I was missing
yeah haha
I was keeping focus on the intersections
and realizing they were finite
and forgetting the actual infinite product in the preimage
thank you
unkempt bolt
You need to show each is not coarser than (not a subset of) the other. In other words, there's a set A which is open in the first but not the second, and a set B which is open in the second but not the first.
nimble cipher
Okay that clears my confusion. You are great!
nimble cipher
How can I prove that the topology generated by $\mathcal{B}={[a,b):a,b\in\mathbb{Q}}$ is different from the lower limit topology?
gentle ospreyBOT
emphatic_wax:
nimble cipher
Is this going to be helpful?
unkempt bolt
Yeah that lemma generally helps. Usually when I think of finer, (2) is what I think of
Do you know what you have to do?
coral pawn
sleek thicket
yes you can 
rugged swan
Check what means sx=tx
sleek thicket
it got solved in algebra
honest narwhal
Alright @sleek thicket and @pastel linden check this
gentle ospreyBOT
Daminark:
honest narwhal
No you must learn what real difftop is you're gonna love it so much
Just bear with me
pastel linden
differential geometry/topology is just objectively more fun than analysis
honest narwhal
So yeah let's say the two manifolds are of the same dimension
gentle ospreyBOT
Daminark:
sleek thicket
yes? This is the same as the degree as a map on top de rham cohomology
honest narwhal
And now there's a more general notion of this form of degree that I wanna introduce, I want to talk about it in terms of point counting first but eventually you'll get to cohomology
Or just in case you know of it already in Lee, are you familiar with transversality?
sleek thicket
yes
honest narwhal
I didn't think Lee did that, so okay nvm I was gonna talk about intersection numbers and cup products
sleek thicket
this stuff so far was all covered in Lee (the thing about maps being classified by degree was stated but not proved)
pastel linden
I'm only just starting manifolds in Lee I'm lost
sleek thicket
I think we talked about that in my AT reading group
Isn't the analogue of the cup product for de rham the wedge?
honest narwhal
Yeah
sleek thicket
I thought intersection numbers were about the cap product
honest narwhal
Uh, so the punchline is that if you give me two submanifolds of a manifold, you push over their fundamental classes
Take the Poincare dual, you get cohomology classes
sleek thicket
yeah
honest narwhal
Take the cup product
Dualize that guy
That's the fundamental class of the intersection
sleek thicket
neat
honest narwhal
But yeah for the proof of Hopf degree theorem, you need a technical lemma
That given two points in a connected manifold, there's a diffeo sending one to the other which is isotopic to the identity
sleek thicket
Can't you do this with flows?
Like I feel like I could give a proof of this right now
honest narwhal
Yeah pretty much
sleek thicket
it's local
honest narwhal
You do it on a ball
sleek thicket
Yeah exactly
Put one at the origin
Try to use scaling
Ono it needs to be compactly supported
Like be the identity outside of some smaller ball
okay but we can patch this by recognizing scaling as the flow of a vector field
Make that vector field compactly supported
but do so without fucking up its flow for some fixed time in a compact region
which gives us a flow sending one to the other which is the identity outside a ball
The flow itself gives an isotopy between the identity and the flow at time 1 we're using to move one point to the other
that's a proof right?
honest narwhal
That's basically how I remember Milnor doing it
sleek thicket
We had a homework problem which wasn't quite this lemma
But everyone's solutions were this lemma
like they were obviously isotopic to the identity
honest narwhal
But yeah once you have this you can play some games and get Hopf degree theorem in the smooth case
And then smooth approximation finishes the job
It takes work for sure to get Hopf, I don't remember the deets but they're in Guillemin-Pollack
sleek thicket
Yeah I don't really see why this is helpful
You can take your two maps
Compute the degree relative to some point in S^n
move around the points in the preimage so they're the same
But I don't get why that's helpful
Β―\_(γ)_/Β―
maybe I'll just look at that book
honest narwhal
Okay so
sleek thicket
im trying to think about analysis
Intuitively two things can be finitely wiggly but their product is infinitely wiggly
But I can't think of an exampl
honest narwhal
The idea is basically that if you have a map f:R^n->R^n such that f^{-1}(0) is finite
And the signed count is 0
Then you can find some g:R^n->R^n\{0}
Which equals f outside of a compact set
sleek thicket
seems plausible
honest narwhal
Okay checking GP again I confused things, what I was thinking about that lemma was that it let you use Hopf theorem to show that a manifold has a nowhere vanishing vector field iff its Euler char is 0
And you want to say that every compact manifold has a vector field with finitely many zeroes, and in fact if you give me an open set you can choose them to lie in it. Being able to make them lie in an open set is where you use an extension of this lemma
Because the idea is you just say okay they are where they are, push them into an open set by a diffeo
sleek thicket
Ah I see
honest narwhal
So you just say okay do it in an open set
sleek thicket
I think I see maybe how you'd go about doing finitely many zeroes?
honest narwhal
And then play the game in R^n
sleek thicket
Yeah exactly
Cover by charts
Use compactness
Don't screw far away things up
In a chart
knotty pasture
I have a problem of differential geometry/algebraic topologuy
A very little problem
Who know the Cartan's Magic Formula?
I mean $\mathcal{L}_x=\iota_x\circ d +d\circ\iota_x$
gentle ospreyBOT
Davide:
knotty pasture
Obliviously this formula have sense only for k-form
My answer is: $\mathcal{L}$ is a derivation operator i.e $\mathcal{L}(\omega\wedge\nu)=\mathcal{L}(\omega)\wedge\nu+\mathcal{L}(\nu)\wedge\omega$. But $\iota_x$ and $d$ is antiderivation operator
gentle ospreyBOT
Davide:
knotty pasture
my book say that since $\iota_x$ and $d$ is antiderivation operator then $\iota_x\circ d$ is derivation operator. Why???
gentle ospreyBOT
Davide:
gentle ospreyBOT
shamrock:
sleek thicket
What is $f(g(\omega \wedge \eta))$
gentle ospreyBOT
shamrock:
sleek thicket
I just expanded everything and got $f(g(\omega)) \wedge \eta + \omega \wedge f(g(\eta))$
gentle ospreyBOT
shamrock:
sleek thicket
You do end up with cancelation
What was your final answer?
Wait no sorry I misread something
I am now just as confused as you
knotty pasture
xD
sleek thicket
I end up with an extra $(-1)^k (g(Ο) \wedge f(Ξ·) + f(Ο) \wedge g(Ξ·)) $ term
gentle ospreyBOT
shamrock:
knotty pasture
I end up with an extra $(-1)^k (g(Ο) \wedge f(Ξ·) + f(Ο) \wedge g(Ξ·)) $ term
@sleek thicket YEs
gentle ospreyBOT
Davide:
knotty pasture
and also if you sum $d\circ\iota_X+\iota_x\circ d$ you don't obtain cancellation
gentle ospreyBOT
Davide:
sleek thicket
hmmm
knotty pasture
but for some funny reason
for this it works $d\circ\iota_X-\iota_x\circ d$
gentle ospreyBOT
Davide:
knotty pasture
other thing don't work for this obliviously
sleek thicket
The degree of g(w) is not k
It's k-1
Or k+1
For the interior multiplication/exterior derivative
Err sorry that's not quite right either
No yeah I agree with myself
You end up with an extra $(-1)^k (f(Ο) \wedge g(Ξ·)) - g(Ο) \wedge f(Ξ·) $ term
gentle ospreyBOT
shamrock:
knotty pasture
i try now
sleek thicket
Oh yeah I think I see it
The roles of f and g will swap
When you reverse the order of d/ΞΉ_X
And so the signs of f(Ο) ^ g(Ξ·) and g(Ο) ^ f(Ξ·) will swap
And then they'll cancel
Let me know if that works for you
pseudo crane
Can someone hell me come up with the coordinate expression for the 1-form g(V), where V is a vector field and g is a metric space isomorphism from T to T*?
In this case T=R^3
g is defined by g(v)=inner product of v and w
sleek thicket
I think if you write V with respect to an orthonormal frame g(V) should have the same expression but with respect to the dual coframe
So in this case our orthonormal frame is just d/dx, d/dy, d/dz
And V = a d/dx + b d/dy + c d/dz
sleek thicket
if W = p d/dx + q d/dy + r d/dy then g(V)(W) = a p + b q + c r = a dx(W) + b dy(W) + c dz(W) = (a dx + b dy + c dz)(W)
oh
A frame is a smoothly varying basis for a tangent space
It's like a basis on the space of vector fields
pseudo crane
Huh
sleek thicket
Instead of a basis for the tangent space at a single point
Sorry I misjudged what level of abstraction you were asking at
The g threw me off lol, I've only seen that for riemannian manifold stuff
pseudo crane
This is manifold stuff
Not riemannian manifold
Let me look at what you've written for a sec
sleek thicket
okay so do you know d/dx and dx are?
In my notation
The d in d/dx should be a partial symbol
pseudo crane
Is d/dx equivalent to partial/partial x?
sleek thicket
yup
Also a, b, c, p, q, r are smooth functions R^3 -> R
I'm assuming your manifold is R^3, but maybe I've misunderstood what you're asking
pseudo crane
Yes, T=R^3
sleek thicket
Isn't T your tangent space?
pseudo crane
Uh
Let me check
My book considers the possibly n-dimensional vector space T to be R^3 for this
sleek thicket
I'm a little confused because you've said V is a vector field
And g is an isomorphism from T to T*
But you're plugging V into g
pseudo crane
X?
sleek thicket
Sorry, V
So I assumed T was the tangent space of some manifold M and V a vector field on M
pseudo crane
I think they are identifying T(R^n,x) with R^n for each x\in R^n
sleek thicket
Okay, this was my understanding
I just wanted to make sure
Since I got a little confused
Then everything I've said should be valid
pseudo crane
What I'm actually trying to do is calculate a coordinate expression for curl V
sleek thicket
If V = Ξ£_i V^i d/x^i then g(V) = Ξ£_i V^i dx^i
does that make sense?
And the proof in the case of R^3 is here:
if W = p d/dx + q d/dy + r d/dy then g(V)(W) = a p + b q + c r = a dx(W) + b dy(W) + c dz(W) = (a dx + b dy + c dz)(W)
@sleek thicket
pseudo crane
I don't understand the third step
sleek thicket
When I factor out the dx, dy, dz?
pseudo crane
Going from, for example, ap to a dx(W)
pseudo crane
Exterior differential of the coordinate function x?
sleek thicket
Yeah but I mean
pseudo crane
Er
pseudo crane
Oh sorry I was over complicating. Just W(x)?
sleek thicket
Yup! And what's that?
Yeah it's important to remember that the exterior derivative of a function (0-form) is easy to compute
Even if the exterior derivative of a higher degree form is nasty
If W = p d/dx + q d/dy + r d/dz, what is Wx?
pseudo crane
Do I need the definition of the coordinate function for this?
sleek thicket
I guess?
pseudo crane
Like using the coordinate systems on the manifold X
sleek thicket
You need to know how the differentials d/dx, d/dy, d/dz act on the coordinate functions
But this is just calculus
$\frac{\partial x^i}{\partial x^j} = \delta^i_j$
Right?
gentle ospreyBOT
shamrock:
sleek thicket
So what's Wx?
pseudo crane
p?
sleek thicket
Yeah
pseudo crane
Ah I see
sleek thicket
More generally, if $W = W^1 \frac{\partial}{\partial x^1} + \ldots + W^n \frac{\partial}{\partial x^n}$ in local coordinates, then $dx^i(W) = W^i$
gentle ospreyBOT
shamrock:
sleek thicket
Does that make sense?
pseudo crane
Yes
I knew that you had to replace the del/del x with dx but I couldn't explain why
sleek thicket
the covector fields dx^1,...,dx^n are super easy to write down and form a coframe ("smoothly varying basis") for the cotangent bundle over a coordinate chart
which is good
yeah I mean it just sort of feels right
pseudo crane
basis for TX is to basis for T*X as frame is to coframe?
sleek thicket
Yup
pseudo crane
Ok
sleek thicket
Coframe is a bad term I shouldn't use
it's also a frame, there's no reason to have special terminology for T*X
pseudo crane
So my book specifies curl V as m(d(g))(V)
sleek thicket
what's m?
pseudo crane
Defined by $m(\phi)= \lambda$, where lambda satisfies $\phi \land \psi =\lambda\omega$ for some volume element omega
Oops
sleek thicket
No worries
gentle ospreyBOT
IAmJon:
pseudo crane
Can't remember what the symbol for exterior power is called
sleek thicket
Is m a map from Ξ^(n-1) T^* -> T^*?
Okay yes I think I agree with everything now
I'm on board
pseudo crane
Ξ^(n-1) T^* -> T
pseudo crane
phi is in Ξ^(n-1) T^*
pseudo crane
Ok
sleek thicket
Oh okay sorry
The important thing is that you can define a vector by its action on covectors
Which is what's happening here
With m
sleek thicket
So m satisfies Ο ^ Ο = Ο(m(Ο)) Ο where Ο is the volume form and Ο any covector, Ο an (n-1)-form
and this uniquely specifies m
Okay so curl V is m(d(g))(V)?
I'm confused again lol
pseudo crane
Yeah, I got as far as writing the coordinate expression for g(V), even if I couldn't justify it
sleek thicket
g is a 2-tensor, right? But it's not alternating so how are we taking d of it?
pseudo crane
But when I take the differential, since it's linear, wouldn't it distribute to the dxi?
sleek thicket
I don't understand the expression m(d(g))(V)
pseudo crane
My book says g(V) is a 1-form, which makes sense
sleek thicket
Yeah I am on board now
sleek thicket
On a 3 manifold n - 1 = 2, so we can apply m to d(g(V)), which is a 2 form
pseudo crane
Yep
pseudo crane
Wait shouldn't we compute the expression for d(g(V))?
pseudo crane
But when I take the differential, since it's linear, wouldn't it distribute to the dxi?
sleek thicket
So choose local coordinates
Let V = a d/dx + b d/dy + c d/dz
Then g(V) = a dx + b dy + c dz
And so d(g(V)) is uhhh God i hate computing exterior derivatives
pseudo crane
It's linear, isnt it?
pseudo crane
Oh
sleek thicket
d(a dx) = da ^ dx I think? Maybe there should be a minus sign?
I think there's no sign
gritty widget
that looks right
sleek thicket
Okay so d(g(V)) = da ^ dx + db ^ dy + dc ^ dz
pseudo crane
At the top there
sleek thicket
yup
And in this case ΞΌ = a is a 0 form
so k = 0
The other term in the sum vanishes because it contains d(dx)
pseudo crane
Oh right
pseudo crane
Ok I might be able to do it from here
sleek thicket
I didn't see him coming
I was terrified
And he got his dirty paws on my analysis book
adorable
pseudo crane
lol
pseudo crane
So you said lambda = psi(m(phi))?
sleek thicket
Yes
when it says m(phi)(psi) it really means psi(m(phi)) lol
oh I see how to do this
pseudo crane
Oh that's confusing
pseudo crane
^ is distributive in the grassmann algebra right?
pseudo crane
Wdym by that
sleek thicket
Actually it's bilinear over smooth functions too
I mean it's a bilinear function of its arguments? Linear in each when you hold the other fixed
pseudo crane
Oh ok
sleek thicket
(aΟ+bΞ·)^Ξ½ = a(Ο^Ξ½) + b(Ξ·^Ξ½)
I originally said this only holds for a, b scalars but it also works if they're smooth functions
gentle ospreyBOT
IAmJon:
pseudo crane
Oh right
That makes more sense
I've got to do something, but I'll be back in a second
pseudo crane
Ok I'm kind of piecing it together
The problem is the definition of the wedge product this book gives is super weird
Is this standard?
sleek thicket
That's not the books fault
The wedge is just weird af
You shouldn't use the definition
Just use properties like antisymmetry/bilinearity/etc
I've used the definition like exactly once after proving all the standard properties about it
pseudo crane
So da ^ dx ^ psi = (β_x(a)dx + β_y(a)dy + β_z(a)dz) ^ dx ^ (a1 dx + a2 dy + a3 dz) = -β_y(a) a3 dx ^ dy ^ dz + β_z(a) a2 dx ^ dy ^ dz for smooth functions a1, a2, a3?
And likewise for the other terms
Sorry it took me so long; I got confused by trying to use the definition of the wedge product
Does that look right?
sleek thicket
I'm sorry I don't want to check that for correctness right now
It's too messy for me
pseudo crane
That's fine, but is the structure at least what you would expect?
sleek thicket
It should be a linear combination of wedges of dx, dy, dz but beyond that I can't say
pseudo crane
I kinda don't want to do the rest because it's a little messy but I'm pretty sure this is the right method based on what I know about the curl of a vector field
Thanks for all your help! This book is a little bare bones so it's really useful to talk to someone who knows about the subject
slim surge
I saw this question, and Iβm thoroughly stumped: βSuppose f is a smooth, compactly supported function on R^2. If you know the integral of f over every line in the plane, can you determine f?β
sleek thicket
does line mean line and not curve?
Like, straight lines
Not necessarily through the origin?
sleek thicket
And f is scalar valued?
slim surge
Yes
sleek thicket
I think I have a disproof but I'm not sure
Sorry I mean a proof
Suppose $\int_\ell f = \int_\ell g$ for all lines $\ell$. Let $h = f - g$ so $\int_\ell h = 0$ for all lines $\ell$. Suppose $h$ is nonzero, i.e. $f \neq g$. Then there's some point at which $h$ is not zero. In a small ball around that point $h$ is either always strictly negative or strictly positive. Wlog $h > 0$ in the ball. Choose a line $\ell$ in that ball. Then $h$ is a strictly positive on the line but has zero integral over it, a contradiction
gentle ospreyBOT
shamrock:
slim surge
That was similar to my first thought too, I was considering do something with segments
But the line version is much trickier
sleek thicket
Yeah for sure
slim surge
One thing Iβve consider, is that you can find the boundary of the bumps
sleek thicket
If it's only lines around the origin you can make some kind of sinusoid which is nonzero but has zero integral I think
Like the boundary of the support?
slim surge
You can shoot lines in different directions, and boundary of the support is the boundary of where those lines switch from zero to non-zero
At least, that would be valid if the function was either always non-negative or non-positive
I was considering whether itβs possible to construct something that just leaves out some area, like this
It seems like maybe thereβs some way to get the area of the entirety of the same, besides some excluded area. You could then maybe take a limit as that area gets small to find the value at a point
Just a thought, I havenβt been able to prove anything with this, and Iβm also assuming thereβs only one βbumpβ in the whole domain, otherwise Iβd have to deal with those lines hitting another non-zero part
red garden
dr strange portal
slim surge
Oh lol your right it does look exactly like that
cursive flume
@pseudo crane maybe this will help,but not sure
knotty pasture
Hello every one!
Could anyone give me a justification of why every isometry between pseudo-Riemannian manifold send Levi-Civita connection in Levi-Civita connection?
pseudo crane
@cursive flume thanks, my book doesn't give any motivation for the definition so that's helpful
cursive flume
well, I study physics and the main idea how wedge product was motivated to me is volume forms
a mathematician could say this is not super precise,but it's a motivation afterall D:
willow spear
can someone explain to me in simple terms what the finite-closed topolgoy is?
tough imp
Uhhh, guessing from the name you just take all finite subsets, empty set, and entire set as the closed set
This is the same as the cofinite topology where a set is open if its complement is finite
sleek thicket
I think you can do this in two steps
cursive flume
wut
sleek thicket
Let Ο be a k form
Suppose the identity holds for Ο
Show it folds for f Ο for a smooth function f
also clearly the identity holds for sums of k forms when it holds for both since lie derivatives are additive
that's step 1
The next step is that if the identity holds for Ο, it holds for dΟ
Since lie derivatives commute with the exterior derivative
right?
cursive flume
yes i have seen these proofs but the problem is i dont know anything of exterior derivatives yet
this is all I know,and that a k form is a totally antisymmetric (0,k) tensor field
sleek thicket
I'm not sure what you mean by "these proofs"
cursive flume
I have seen like 3 proofs with exterior derivatives in diffgeo boosk
sleek thicket
I'm just trying to show you how to extend from k forms to k+1 forms
cursive flume
can you define the exterior derivative pls?
sleek thicket
and then by induction you'd get it for all forms
no lol
It's very complicated and awful
cursive flume
ah ok
sleek thicket
Okay hmmm
The reason I was to use it is that it takes k forms to (k+1) forms
And the image generates Ξ©^(k+1) if you allow linear combinations with smooth function coefficients
ugh it doesn't even tell you how to compute the lie derivative?
That sucks
honest narwhal
Exterior derivative is fine
cursive flume
but i don't use that definition
honest narwhal
Define it for 1-forms
sleek thicket
wym dami?
cursive flume
sleek thicket
I find it very awkward to define
cursive flume
i derived these formulas by expanding in charts
sleek thicket
you need to do it in charts and make sure things glue
cursive flume
but note,i'm not at any high level of diffgeo,not following a maths book
honest narwhal
Oh I was thinking in R^n
honest narwhal
Well there it's just
cursive flume
i'm following these lectures
sleek thicket
I look it up when I need it
honest narwhal
d(f dx_{i_1} \wedge ... \wedge dx_{i_n}) = df \wedge dx_{i_1} \wedge ... \wedge dx_{i_n}
So just define it for 1-forms
sleek thicket
oh sure
honest narwhal
And boom
cursive flume
i don't know wedge product either π¦
sleek thicket
I don't know why I thought it was bad
honest narwhal
Also, and this is my favorite part
sleek thicket
prophet are those formulas for Y a vector field?
sleek thicket
Yeah it's okay in that case since it coincides with the lie bracket
sleek thicket
but it doesn't tell you about the case for other tensors
cursive flume
sleek thicket
like for example k forms
cursive flume
it does
sleek thicket
it does?
cursive flume
because i can use the leibniz rule
and evaluate the form in a vector field
giving a function
sleek thicket
oh I see
honest narwhal
So let G be the diffeomorphism group of M. It acts on k-forms by pullback, so \Omega^k(M) is actually a G-rep
sleek thicket
that's ugly
cursive flume
i computed it in the above calculations
honest narwhal
One may ask what the G-equivariant maps are \Omega^k(M) -> \Omega^j(M)
cursive flume
yes,it's not fancy,but i don't know more diffgeo,so i gotta do it with what i know π¦
sleek thicket
dami stop
this person had an actual question
And you've already told me this before
honest narwhal
I don't know what the Lie derivative is
sleek thicket
tfw
honest narwhal
Keep in mind my difftop class basically dodged all the technical stuff
We just took a geodesic path to Whitney
sleek thicket
okay so basically like
honest narwhal
And stuck everybody in R^n
cursive flume
i got this answer
but i can't comprehend it since i'm not following a math course π¦
honest narwhal
"what? lmao just use a basis"
Holy fuck
sleek thicket
lol that's a name I haven't seen in a while
sleek thicket
fucking hank
Prophet I'll think about your problem
It's awkward because you have so few tools
cursive flume
yes,basically i have only the def of lie-derivative
sleek thicket
@honest narwhal okay so if v is a vector in R^n
And f a smooth function R^n -> R
how do you compute the directional derivative of f wrt v?
cursive flume
i have done it for 1 forms but i seem to struggle generalizing it π¦
honest narwhal
Do we know Df? Then it's Df(v) lol. Or you take the limit as usual
sleek thicket
Obviously you take the flow ΞΈ of the vector field which is constantly v
Sorry checking details lol
cursive flume
sleek thicket
Memerock imo
cursive flume
isnt it like this,but then the symbol del/del x^i will be the real partial derivative?
cursive flume
i know what a flow is but just the very basics
sleek thicket
sorry I'm explaining something to dami
cursive flume
so i have integral curves and complete vector fields
sleek thicket
Okay I got my detail right
cursive flume
a flow is a one parameter family Rxm->M which maps the point and the paramater to h_x^ lambda at the point p := the unique integral curve at the point
aah ok
sleek thicket
@honest narwhal so the actual directional derivative is lim (f(x+hv) - f(x))/h, right?
honest narwhal
Yee
sleek thicket
but what if f is not a function on R^n but a function on M
Or a vector field
Or a tensor field
suppose X is some tensor field on M
sleek thicket
Then (X(x+h) - X(x))/h isn't defined
cursive flume
exactly π
cursive flume
you need to push forward to be in the same tangent space
honest narwhal
Stolen
sleek thicket
basically the lie derivative is like a directional derivative but because we're on a manifold we need a vector field defined near the point and not just a single vector
that's the punchline
In R^n this is the straight line/constant vector field of your direction vector
cursive flume
do you have a simple explanation of how to get the lie group of a lie algebra?
cursive flume
as i stated,just finished undergrad physics,and i derived the isometries of minkowski spacetime
and got the poincare algebra
sleek thicket
There's a unique simply connected lie group with a given lie algebra
cursive flume
now i'd like to reconsruct the group
sleek thicket
I don't know what any of that means lol
cursive flume
isometry=distance preserving diffeomorphism
sleek thicket
okay I knew that part
That's the one word I knew
I also knew "undergrad" and "physics"
cursive flume
sleek thicket
this means nothing to me still
I don't know enough riemannian geometry and/or physics for this, sorry
cursive flume
ah ok,np
sleek thicket
but in general two nonisomorphic lie groups can have the same lie algebra
Like
You know how a lie algebra is local to the identity of a group?
cursive flume
i heard something like i could reconstruct the lie group around the identity or something like that
sleek thicket
It's sort of the tangent space at the identity
Yeah you can do that
but only in some small neighborhood of the identity
cursive flume
yes
sleek thicket
This is what the exponential map does
sleek thicket
Yes, that's the defintion of the exponential math
It's smooth and a local diffeomorphism
cursive flume
if so,then i think it is recommended to watch the lecture
cursive flume
it's from this
i just finished his lectures on gr and started to do this,i'm at lecture 5 tho. the topological invariants seem very abstract to me π
like the stuff with fundamental group
also he names like 5 theorems he doesnt prove and says do it as homework,so i struggle with those
sleek thicket
The fundamental group is very based
Okay so I should clarify
We have approximately opposite backgrounds here
It sounds like
I do not know any physics
But I am comfy with smooth manifolds and topology
cursive flume
thing is i dont see the point of introducing the topological invariants and stuff like that
for me as a physicist it is very abstract
why would i like to classify topological spaces?
sleek thicket
Yeah I couldn't tell you why you'd like to
cursive flume
topology is indeed super important,in order to be able to tlak about continuity
and then the paracompact condition is mega important,because then the topological spacea dmits a partition of unity right
and we need that for integration on manifolds
but the rest just seem black magic to me
sleek thicket
What's the rest?
cursive flume
compact, heine borel theorem
cursive flume
sleek thicket
compactness arguments are extremely common
sleek thicket
anyways I couldn't tell you why a physicist would care about any of this
Because I don't know why physicists care about anything
cursive flume
and connectedness/path-connectedness
prove things about manifolds like?
most of the things i needed as a physicist,didn't require these
liek construction of tangent space into a vector space,from the definition of velocity being a linear map taking me from the smooth functions o nthe mfd into the real numbers
bundle,etc.
i didn't use any of those properties there
where would i use them?
yes,true,that's a good motivation for paracompactness
and you need partition of unity to integrate
are these things needed to prove things in surgery theory/morse-radon theorems?
the lecturer said that those are interesting mathematical areas,but didn't go into any details since he said they're advanced
cursive flume
sloppy physicist :lul:
btw does anyone know how to get from line 5 to 6?
lecturer said the problem is that the object is ill defined since you do the derivative of the tangent vector in all possible directions,but this does not make sense,because the vector field isn't everywhere,it is only along the curve
however it is immediately projected by the \dot \gamma^m into the direction
idk he said this is not mathematically precise
shadow charm
fiber bundles are confusing... Ill probably get a better understanding after some examples but Im having trouble grasping the idea of the structure group
uncut geyser
@cursive flume along the curve just means the application points are in the curve. directions are still "unrestricted"
frosty sundial
@fading vale just because the other channel seems to be occupied now
fading vale
yea
frosty sundial
hmm yes something does seem fishy here
it seems like he's saying that adding q/2 is the same as taking a negative in R/Z
frosty sundial
wait did I say somethign wrong?
frosty sundial
oh
that is strange
but also I think I understand now
-h(s) is the point on the circle directly across from h(s)
thinking of S^1 as R/Z,
going half way around the circle is the same as adding 1/2
fading vale
ohhh
frosty sundial
(or 3/2 or 5/2 or)
frosty sundial
I was a little confused because the - in -h(s) isn't like the group operation on S^1
it's like - in R^2
np :D
limpid mural
Anyone familiar with parallel transport here?
I am having hard time to see how it extends to tensors, I can't see how to define it for dual of the tangent space
I think it's the inverse of the transpose of the parallel transport map between tangent spaces
but then how do I compute it? Seems a bit messy
marsh forge
By parallel transport you mean something of the form L(v)=v+w
Then the obvious (to me) way to transport it would be L(v*)=v*+w* up to some choice of basis
limpid mural
the background is a bit more complicated, we have a manifold $M$ with a connection $\nabla$ on the tangent bundle. The connection induces the covariant derivative $D_t$ on vector fields $X$ along a smooth curve $\gamma$. We say that a vector field $X$ along $\gamma$(that means $X(t) \in T_{\gamma(t)}M$) is parallel if $D_t X = 0$. Now turns out that the equation $D_tX = 0$ is a linear system of differential equations, so I can define a map $\Gamma(\gamma){t_0}^{t} : T{\gamma(t_0)}M \to T_{\gamma(t)}M$ in the following way: to each vector $v \in T_{\gamma(t_0)}$ we define $\Gamma(\gamma)_{t_0}^{t}(v) = X(t)$ where $X$ is the unique parallel vector field along $\gamma$ such that $X(t_0) = v$
gentle ospreyBOT
emme:
limpid mural
$\Gamma(\gamma)_{t_0}^{t}$ is an isomorphism between tangent spaces, now I want to extend it to the duals. I think it could be done in this way
gentle ospreyBOT
emme:
limpid mural
$\Gamma(\gamma){t_0}^{t}^{\star} : T{\gamma(t_0)}M^{\star} \to T_{\gamma(t)}M^{\star}$ sends $\phi \in T_{\gamma(t_0)}M^{\star}$ to the functional $\Gamma(\gamma){t_0}^{t}^{\star}(\phi)$ defined as: $$ \forall w \in T{\gamma(t)}M \ \ \ \Gamma(\gamma){t_0}^{t}^{\star}(\phi)(w) = \phi(\Gamma(\gamma){t}^{t_0}(w))$$
gentle ospreyBOT
emme:
Compile Error! Click the 
reaction for details. (You may edit your message)
supple cipher
Can someone explain to me how to find the first fundemental form of a surface?
limpid mural
is the surface in R^3?
supple cipher
limpid mural
Ok the First fundamental form is the restriction of the standard for product to the tangent planes to the surface
Compute the ta gent vectors and do the standard dot product between them
You'll get the matrix associated to the first fundamental form
unkempt bolt
Just a question, if I might ask. I'm not familiar with the terminology, but from what you said is the first fundamental form the induced metric tensor from the ambient space?
limpid mural
Yes
You restrict the metric tensor from the tangent space to the tangent space of the submanifold that is a subspace of the tangent of the ambient manifold
unkempt bolt
Awesome, that makes sense
Also @limpid mural your map seems good to me, except as you wrote it the domain and codomain of the dual map are switched . I believe it should be ${\Gamma(\gamma){t_0}^{t}}^{\star} : T{\gamma(t)}M^{\star} \to T_{\gamma(t_0)}M^{\star}$ as you wrote it. That said since $\Gamma(\gamma){t_0}^{t}$ is invertible, you can either take the dual of its inverse or take the inverse of this dual map to get a map from $T{\gamma(t_0)}M^{\star}$ to $T_{\gamma(t)}M^{\star}$ (which should turn out to be the same).
gentle ospreyBOT
The_Vman: