but ugh

[what you want them to see](the.url.com)

olivia just get some very bad news.

No, no, I know how to link, silly.

Munchkin is back in full mobility?

I meant an answer on a post here.

As opposed to the post itself.

her leg is stiff, and she says she wants the cast back. why? "i don't know. i just like having it"

Physical therapy?

Tell her the ducks missed her.

we saw a goose with a hurt leg the other week at the duck pond. she was obsessed with it for a while.

Empathy

another attempt

we want functions $f_1,f_2$ mapping $[-1,1]\to [0,1]$ such that...

1) $\int_{-1}^1 f_1(x)\,dx=\int_{-1}^1 f_2(x)\,dx=1$, so these are probability distributions on $[-1,1]$

2) $$\frac{1}{4\pi}\int_{S^2} f_1(\cos\theta)f_2(\cos\theta \cos\alpha+\cos\phi\sin\theta\sin\alpha)\,\sin\theta \,d\theta\,d\phi = \frac14(1-p\cos\alpha)$$

which...yeah

the buddha teaches that wanting things is the root of all suffering

we can joyfully participate in the sorrow of life without attaching ourselves to transient distractions such as $f_1$ and $f_2$

@robjohn I forgot to tag you earlier. I meant linking specifically to an answer on one of the posts.

it looks slightly less weird when you recognize that the second argument is just $\cos\gamma$ where $\gamma$ is the angle between $(\sin\alpha,0,\cos \alpha)$ and $(\cos\phi\sin\theta,\sin\phi\sin\theta,\cos\theta)$

slightly

have you tried pulling this back through spherical coordinates

Coordinates make it worse, Semiclassic.

yeah

Huh? @Thor

i was trying to aim at getting rid of $\theta$ in favor of $z=\cos\theta$

but it really doesn't help

badly worded, I meant to undo the coordinates

yeah

That’s where he started.

I suspect it is about spherical harmonics, as you said.

in the $p=1/2$ case there's apparently an explicit construction

or at least that's how i interpret the source problem

Ask the author?

prof has emailed them

we'll see if they respond

Good

something something expand $f_1,f_2$ in real spherical harmonics

it's not too bad to represent $f_1$, because you can choose its unit vector as the z-axis

and then expand $f_1(\cos\theta)$ in Legendre polynomials

but then the problem is dealing with $f_2(\hat{x}\cdot \hat{b})$

yeah, no better sign of things getting not too bad when you're breaking out the legendre polynomials.

hah

an event of similar status as uncorking bottles of champagne.

relative to having to use full spherical harmonics? yes, yes it is

what about harmonic polynomials? is there room for them?

if i could remember how to use them, yes

i'm forgetting how real spherical harmonics work in that formalism tho

Is munchkin mobilizing?

she's sitting on the couch with a tablet and spelling words like CAT and HAT. no copyright or trademark infringement though. the cat is not in the hat.

she often draws the first letter in her name, over and over, whenever you give her a pen. she has also been known to build it out of french fries. this is going to be a bad signature to have if she wants to go into a life of crime.

Is she supposed to perambulate?

we weren't given any intermediate tools, like an air cast or wrap or crutch or anything, so yeah, i think so. apparently the leg will be stiff for a while (you don't say).

But no PT to rehabilitate?

Just send Olivia to chase her?

yes, we'll pit them against one another. olivia is very good at sensing weakness.

oh, you're walking up the stairs with a cup of hot tea in each hand? time to attack the legs. that kind of thing.

@Xnero Thanks so much! So obvious, but I never looked carefully :P

hmm

i think choosing $f_1(\hat{x}\cdot\hat{a})=\frac12(1+c_1 \hat{x}\cdot\hat{a})$ (and similarly for $f_2$) works, but

the threshold value i'm getting is smaller than i expected

@Xnero I am late to the party but: in the happy world of Cartesian / rectangular coordinates, every point in the plane is uniquely represented by an ordered pair $(x,y)$, which (roughly speaking) say "travel $x$ units east, and $y$ units north" (where negative numbers mean "go the other direction" (i.e. west or south)).

This means that we can uniquely describe a complex number in the form $x + iy$.

On the other hand, in polar coordinates, most points do not have a unique representation. I could say "turn 90 degrees counterclockwise so that you are facing north, then walk 10 feet", or I could say "turn 90 degree clockwise so that you are facing south, then walk backwards 10 feet", or I could say "turn completely around 4,328 times counterclockwise, then turn an extra 90 degrees so that you are facing north, then stumble forward 10 feet".

All of these directions describe the same point in polar land, i.e. $(10, \pi/4)$ (or some similar nonsense).

now it's xander in charge of field sobriety

@leslietownes Yeah, but I am working on a second margarita.

I made FAJITAS tonight!

as your attorney, refuse all field tests, make them take blood at the station

ooh

(which I intentionally mispronounciate as fah-DZI-tahs, just to annoy people).

@leslietownes Oh, my brother is now one of my attorneys. He is the one authorizing the PI to look into the felonious biological parents of my daughter. You know, the biological parents that my sociopathic ex-wife seems to believe are more important to our daughter's upbringing than, you know, her actual father.

is it a long i sound in dzi? you could make it sound like a latin motto

caritas, veritas, fajitas

YES!

one of my regrets thus far as an attorney is that i have yet to have to work with a private investigator

@leslietownes My PI was a friend of my father's---he's doing work pro bono!

sorry to hear somewhat indirectly about an awkward custody situation but happy to hear of that

I really don't want to end up in further litigation, but the way the ex is acting, it seems likely. It is going to be fun to walk into court with an attorney that I don't have to pay. "Hey, lady... if you want to litigate, I can do this all day..."

@leslietownes Yeah, it is kinda crazy.

also sorry to hear of people working for free, can't have that

but i'll allow it

I offered to pay my brother.

you could pay me instead. that way, at least one attorney gets paid

He can't actually accept money from me, because of whatever terms of employment he has with his employer (the public defenders office).

time to unionize

that makes sense. i can't work for anybody either, money or no money.

comrades, we can no longer withstand, tolerate such effrontery

i think i even said i wouldn't represent myself in court. i signed away a lot of things.

But he is being super helpful, and I have been able to invoke some magical words... something like "pro hoc vice".

is this at least local? everyone i've known to be involved in a dispute, it's always spread across states and time zones and everything is more complicated and expensive.

pro hac makes it sound like it's not local

@leslietownes Oh, no. That is part of the problem.

She is in CA, I am in AZ.

The ex does not seem to understand that, in November, when she sets her clock back, I don't.

I would like to schedule calls with my daughter, so my availability changes in November (relative to hers). This is clearly just evidence of me being difficult.

Speaking of which the small child is supposed to be on the phone in 10 minutes. I am excited to tell her about her birthday present, which involves explosives.

yeah, it's you and not the arizona legislature. you control time now.

Every 10 year old needs explosives, right?

enjoy and good luck.

yes.

ty

g'night

Wait. Munchkin will want them.

we gave my daughter a toolkit that contains a functioning hacksaw with multiple blades you can swap in and out.

Surely you jest.

it's not the main feature, it also has a hammer, screwdriver, some bolts. she has a toy set made of wood, and now for some reason a real one. they're small versions of real tools.

Oy.

i don't know why it exists. i'm not saying we would let her use it. yet. seems more like a "wait until 3" kind of thing.

I have a question: Is there any general method to finding the measure of a subset of the reals of uncountable cardinality?

Age 3?

It might be unmeasurable?

@TedShifrin Yeah, true

Also, tools as toys?

Unless it’s described in a prescriptive or proscriptive way, hell no.

So, you have to have a method of defining the elements in the set. Sure

@TedShifrin oh... https://math.stackexchange.com/a/4268017

just the URL on a line by itself

it's called one-boxing

A helpful method.

chat.meta.stackexchange.com/faq#formatting

But all my knowledge of defining sets kind of ends at the jump between countably defined sets and uncountably defined sets. Like, I can algorithmically define singleton sets and add them to the overall set, but I have no intuition for the behavior of this when I'm doing it uncountably many times

I’m lost.

@robjohn If that’s for me, Xnero already told me :)

@TedShifrin oh, about one-boxing. Sorry for the repetition.

One-boxing? Huh?

@TedShifrin that's what it's called when you post a link and the beginning of the link shows up here.

LOL … I am on another planet, but Xnero informed me.

@TedShifrin Yeah, me too

yeah, tools as toys.

Timothy Snyder On Tyranny … read it.

nationalist propaganda smh

@Xnero we could skip the third line if we use this result.

hi ted

No, @shin, anti-nationalist. Shake your head some more.

Hi, fin.

im sober today

Yippee

Calculus speed run

actually, is there any point in doing these in college? mostly it seems introductory

but i may be wrong, i havent watched the full 12 hours of it

hello

I'd be up for chatting number theory and modular arithmetic if anyone might be interested.

I have questions.

for example, I'm trying to understand what modular arithmetic says about the natural numbers and I'm attempting to understand why modular arithmetic and congruence is so important for number theory.

I'm trying to understand conceptually what modular arithmetic says about the natural numbers.

anyone know what ahlfors means by using greens theorem to get (2-1) here? math.stackexchange.com/questions/4270106/…

(This is page 25 of Conformal Invariants: Topics in Geometric Function Theory - Ahlfors)

My journey into Spivak's calculus continues. May I ask what property of numbers allows this construct...used quite extensively in Chapter 1. $ ({ab})^{2} = {a}^{2}{b}^{2}$ Not really explained. Thanks

Spivak's calculus is challenging!

I tried working through the text, but I'm not quite ready for it.

also, TIMTOWTDI

In wikipedia, it explains the method of acyclic models as something related to chain homotopy and chain map. But in Bredon's topology and geometry, it explains something related to extension (implicitly). Does the Wikipedia's explanation of acyclic model cover Bredon's? I can't see how Bredon's is a special case of general method of acyclic model

@zacts Well, I did an intro to calculus during the pandemic with David Easdown from Univ of Sydney as a MOOC. Always wanted to do it and it was inspirational. So, want to go further. Not easy, by any stretch.

@user1115542 that's cool

@user1115542 I don't have Spivak in front of me, but this (1) depends on what you have defined $a^2$ to mean, but (2) should follow fairly immediately from that definition.

@XanderHenderson I have not seen him define exponents. Maybe in the next chapter. 'but I could be wrong

Assuming that $a^2$ shorthand for repeated multiplication (which is an acceptable assumption when the exponent is a natural number), you have $$(ab)^2 = (ab)(ab) = a((ba)b) = a((ab)b) = (aa)(bb) = a^2 b^2. $$ This computation is justified by repeated application of the associative property of multiplication, and an application of the commutative property of multiplication.

That being said, this is a basic result about exponentiation, and it is quite likely that Spivak is assuming that you already know how to work with real numbers (or, at least, rational numbers).

Like... this is something you were supposed to have learned in high school.

@XanderHenderson I like it, and in fact, I think that is covered if you accept his (P7)... which says for every $a$ there is a number $a^{-1}$ such that $a \cdot a^{-1} = a$ Now he does not explicitly talk about the exponent as such, but it's probably pretty close. :-)

@user1115542 You should be careful about that.

$a^{-1}$ is just notation for the multiplicative inverse.

Spivak could just as easily have written $1/a$ to represent that number, or invented some new notation, like, I don't know... $\div a$ (analogous to how the additive inverse of a number is denoted by $-a$, so that $b+(-a) = b-a$).

@XanderHenderson Thanks. Good point

It turns out that this notation is "compatible" with exponentiation, but that is a result which does require some justification at some moment in the development of the theory.

Oh, I just noticed, there is a typo above:

@XanderHenderson At any rate, thanks for that nice proof, which as you point out, does use the associative property.

@user1115542 $a\cdot a^{-1} = 1$, not $a$.

@XanderHenderson Of course. Face plant!!! Thanks Too early in the morning??? :-)

@user1115542 I mean, the sun is up (wonderful sunrise this morning during my commute).

I am about halfway through my coffee for the morning.

@XanderHenderson You must be east of me?

@user1115542 If you are in the US, you would have to be in California to be west of me.

@XanderHenderson Yep...Seattle actually. You know, where it rains all the time...as we try and fool Californians of this fact.

Or Washington or Oregon. :D

(or Nevada or Idaho, I suppose, but I think Idaho is on Mountain Daylight Time, so an hour ahead of me).

@XanderHenderson I just want you to know that for someone like me having these little insights is really inspiring. So thank you.

@user1115542 multiplication without the associative property would be hard, indeed.

@robjohn That's why nobody likes the octonions.

@XanderHenderson I have never looked at them. Now I probably won't :-)

wow, they are hard (not to mention not unique, ouch).

Hi all, I'm having trouble understanding something.

I'm doing Calc III, and I was practicing Linear Approximations

I took the derivative of $ V = \frac{4}{3} \pi r^3 $ and got $ V' = \frac{4}{3} \pi *3r^2 $

Now, this is no big deal, as you might imagine.

But then I saw the books notation:

$dV = 4\pi r^2 dr$

The derivative is taken with respect to $r$?

i.e. $V' = \mathrm{d}V/\mathrm{d}r$?

@XanderHenderson Yes, it was.

Now, I'm trying to intuitively grasp what's really going on here. For too long, I've been throwing around signs and symbols without understanding.

Here's what I understand: $dV/dr$ is the rate of change of the volume with respect to the radius. That means we poke the radius a bit and see how the volume responds. No surprise there.

I think that the best way to think about the differentials (in the context of introductory calculus) is as errors or estimates of errors.

@XanderHenderson Hmm, I'll touch on that. I've also come to understand that $dV/dr$ encapsulates the language of limits within it. So, for instance, $dV/dr$ is an abstraction that really means the limit of the change in volume as the change in radius approaches 0. Am I correct?

so... I'm trying to understand modular arithmetic, congruence, and "residue classes" a bit this AM

The differential $\mathrm{d}r$ is the error in the measurement of the radius of your ball. So you have a ruler which measures in millimeters, so when you say that your sphere is 10.3 cm, you really mean that it is 10.3±0.05 cm. $\mathrm{d}r$ somehow captures that error.

within the context of elementary number theory perhaps

also, don't want to interfere with a current more on-topic discussion. :-)

@XanderHenderson That's an interesting perspective.

You then say "Well, the relation between radius and volume is approximately linear when the change in the radius is small," so you can approximate the resulting error in your computation of the volume by using the local linear approximation.

the first question I have though this AM is: why is divisibility so important in elementary number theory?

"If my error in measurement is not too large, then the resulting error in the estimate of the volume should be about $V'(r) \,\mathrm{d}r$."

Right, as we zoom into our function, it becomes arbitrarily straight. That's the main idea behind Linear Approximations.

Which is the differential $\mathrm{d}V$.

@rb3652 Exactly.

It also becomes the main idea behind differentiation on manifolds (though Dr Shifrin is much more expert on that than I).

I'm kind of looking at divisibility and residue classes as being analogous to the atomic weight kind of for the natural numbers?

I'm confused -- V'(r) dr? $\frac{dV}{dr} * dr = dV$, no?

so I'm not quite so interested in how to solve problems or do computations with modular arithmetic yet, I'm just trying to understand this tool of congruence, "residue classes", and modular arithmetic, and what it's useful for.

Don't treat the differentials in the derivative notation like independent objects.

When we write $\mathrm{d}V/\mathrm{d}r$, that is one symbol.

It isn't actually a fraction.

It's hard to pinpoint my confusion, but that's something I've struggled to understand for a long time.

We can abuse it as though it were a fraction, but it isn't, really.

OK, when I see dV/dr, I immediately think "poke r, see V's response"

When I write $V'(r) \,\mathrm{d}r$, note that $V'(r)$ is a number. It is the derivative of $V$ at $r$.

@XanderHenderson oh that's interestingly cool re notation.

In the notation $\mathrm{d}V = V'(r) \,\mathrm{d}r$, the differential $\mathrm{d}r$ is treated like an independent variable, while the differential $\mathrm{d}V$ is an independent variable.

And I can't treat it like a fraction, because it's a function.

what's the reason for dV/dr notation?

if it's treated as one symbol

So $dV = V'(r) dr$ is akin to $y = mx$

that's something that kind of confused me with dy/dx.

@zacts It has to do with the way that Leibniz understood calculus.

@rb3652 Exactly!

ok

@XanderHenderson. This is the first time I've seen this -- or at least the first time I've thought about it this way. I'm trying to wrap my head around it.

OK, so now my book's equation makes sense -- from that perspective.

@zacts Essentially, suppose that $y = f(x)$ (that is, $y$ is a function of $x$). To compute the rate at which $y$ is changing with respect to $x$, you perturb $x$ by a little bit, then watch to see what happens to $y$. If you change $x$ by $\Delta x$, then the corresponding change in $y$ is $\Delta y = f(x+\Delta x) - f(x)$.

Then take a limit as $\Delta x$ goes to zero: $$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{ f(x+\Delta x) - f(x) }{\Delta x} = f'(x) = \frac{\mathrm{d}y}{\mathrm{d}x}. $$

$dV = 4\pi r^2 *dr \rightarrow dV = V'(r) * dr \rightarrow y = m * x$

Before passing to the limit, $\frac{\Delta y}{\Delta x}$ is a fraction (it is a ratio of honest-to-goodeness numbers). After passing through the limit, the result is no longer a real fraction, though it is reasonable in some circumstances to think of it as a fraction of "infinitesimal" quantities, or the "ghosts of departed quantities" ( maa.org/press/periodicals/convergence/… ).

what's d mean there?

why is it d and not delta?

@XanderHenderson So what I'm taking away from this is that $dV$ itself is a function. And it's slope is $V'(r)$ and the dependent variable is $dr$. Change $dr$ a bit, and $dV$ responds.

@zacts It is just part of the notation. If $\Delta x$ represents a finite perturbation, $\mathrm{d}x$ represents an infinitesimal perturbation, I suppose.

oh interesting

@rb3652 Essentially, yes.

I think that is a reasonable way of thinking about it.

@XanderHenderson Just one thing -- $dV$ need not be linear, no? So why is it written in $y=mx$ form? Or does the $V'(r)$ compensate for that?

@rb3652 Ah, but $\mathrm{d}V$ is linear. It is the approximation of the error you get via the local linearization of $V$.

Whoa, whoa. Let me chew on that.

Remember, $\mathrm{d}r$ is the independent variable, and $V'(r)$ is "just" a number.

Because you have fixed $r$.

OK, I'm going to write this out (or I'll forget) and it would be great if you could let me know if I made any mistakes.

(@TedShifrin is likely to slap me around a bit for the following, but I believe that the correct interpretation of this (from a higher level) is that $\mathrm{d}V$ is something like an element of the tangent space over the point $r$.)

(I learned differential topology from a homotopy type theorist, and I have never bothered to try to learn it more gooder, so I am kind of weak there...) :\

Still writing it out.

No rush. Though I am going to have other things to do soon---in need to be online for the first class of the day in 20 minutes.

And I really should be reading these scholarship applications...

@XanderHenderson are you undergrad?

@zacts No. I am faculty.

oh cool

I'm undergrad soon.

@XanderHenderson OK, finally done. Thank you for patiently waiting.

@rb3652 I would say that is close, but not quite right.

@XanderHenderson OK. What can I improve?

$\mathrm{d}f$ is not $f(x+h) - f(x)$.

Rather, it is $L(x+h) - f(x)$, where $L$ is the local linear approximation.

Oh, oh, I see. I see, yup.

I should have drawn a tangent.

@XanderHenderson Perhaps this would be more correct:

Yeah, that seems okay.

And is the limit correct?

The limit is correct, but I am not sure that I would want to think about it that way.

Oh?

What would you suggest?

I mean, it is correct because $f(x) = L(x)$, so you are really just computing $L'(x)$.

But $L$ is the line tangent to $f$ at $x$, so $L'(x) = f'(x)$ by construction.

Right. (Circular Reasoning?)

So it isn't wrong, I suppose, but it has an off flavour.

How would you advise me to think about $\frac{dy}{dx}$?

@rb3652 It is the derivative. How do you want to think about the derivative?

Change.

A little change in input and see how the function responds.

It is an instantaneous rate of change, or the slope of a tangent line.

Right.

reading 30-year old papers is interesting

especially when they're a bit dense

@XanderHenderson I totally understand, it's just that when I saw $dV = 4\pi r^2 * dr$ in my textbook, I was caught off guard, because I realized I didn't really understand what it meant. Now I see it's a linear approximation scenario.

case in point: www2.fisica.unlp.edu.ar/materias/qc/Sp.pdf

the funniest sentence i've seen so far in there:

What they mean: By definition, a density matrix $W$ is normalized to have trace one, so...

what they write: "For density matrices $W$ $\text{tr}W=1$ is fixed, so..."

i get what they meant, but it reads terribly

As per math.stackexchange.com/questions/340744/…, "On their own 𝑑𝑦 and 𝑑𝑥 don't have any meaning", which is why I was confused when I saw that equation in my textbook.

they don't, but we often pretend they do

and often we can get away with it :P

Yes, but that doesn't make any sense. What did make sense to me was @XanderHenderson's Locally Linear Approximation perspective of $df = f'(x) dx$

That made sense because now, $df = f'(x) dx$ is just $y=mx$. It's just a line (locally).

Also $dx,ds$, etc do have meanings in higher-level math; they're defined to act on tangent vectors in a way entirely analogous to how row vectors act on column vectors

but, uh

that's not something you should try to use in calc 1

(it's not something i really use at all, and i employ calculus on a daily basis)

I'm in Calc III, and I've mindlessly thrown around dx and dy's as if they were fractions and whatnot and I've decided now is finally the time to reckon with my misconceptions.

yeah

I think I've got an OK understanding for now. Thank you.

Hello

Where to ask questions about knapsack problems and greedy approaches please?

What tags I should use on mathexchange?

@XanderHenderson Um, no, $dx$ is the dual of a tangent vector, and $dV = dx_1\wedge\dots\wedge dx_n$ is an $n$-form which is the "volume element."

@TedShifrin Hey, I hedged like crazy! But okay, cotangent thingy. That actually makes more sense.

@TedShifrin. Any idea pleace how to trace the trajectory of gradient over 3D surface!

Which topic discusses that please?

Yeah, okay. That makes more sense. $f'(a)$ is a velocity, so $f'(a) \,\mathrm{d}x$ is a momentum. So $\mathrm{d}x$ eats a velocity and spits out a momentum. So it lives in the cotangent bundle. Yum.

i should relearn how tangent/cotangent bundles go with Lagrangian/Hamiltonian dynamics

@Semiclassical That is one of the things that I am struggling to learn this year.

Except that that I am also trying to learn Lagrangian and Hamiltonian dynamics, and all of this is in the framework of span categories. I feel really dumb most days.

But I really have to be off now. Students might show up any minute...

i know it at the level of: Hamilton's equations can be written as $dq/dt=\partial H/\partial p, dp/dt = -\partial H/\partial q$

and therefore $dH = p\,dq-q \,dp$

but that's...not much

@TedShifrin here's something I ran into yesterday. I think i know what it looks like, but i'm not convinced

Suppose $\hat{a},\hat{b}$ are unit vectors. What does the set of unit vectors $\hat{n}$ such that $(\hat{a}\cdot\hat{n})^2>(\hat{b}\cdot\hat{n})^2$ look like?

@Semiclassical would it be a half sphere?

that's what i thought initially but i seem to be getting a quater sphere instead

which, confuses me

without loss of generality, take $\hat{a}=(\cos(\alpha/2),\sin(\alpha/2),0)$ and $\hat{b}=(\cos(\alpha/2),-\sin(\alpha/2),0)$ where $\alpha$ is the angle between $\hat{a},\hat{b}$

then the inequality becomes $$(\cos(\alpha/2)n_x+\sin(\alpha/2)n_y)^2>(\cos(\alpha/2)n_x-\sin(\alpha/2)n_y)^2$$

so $0<2\cos(\alpha/2)\sin(\alpha/2)n_x n_y=n_x n_y \sin\alpha$

since $\sin\alpha>0$ for $\alpha\in [0,\pi]$, this seemingly amounts to $n_x n_y>0$

which, actually, is half a sphere, but

it's like you took two opposite quarter spheres, not a hemisphere

to put it another way, the set where you have equality seems to be the union of two orthogonal great circles

which i find odd

@Semiclassical I have to leave shortly for a while, but surely it depends on $\hat a$ and $\hat b$. But I insist on working geometrically; you like formulas too much. If they're the same or opposite, we get nothing. I think you should work in the plane spanned by the two vectors and then ultimately rotate about the angle bisector. I'll work on it more when I get back.

yeah, i do have a geometric picture. take the great circle connecting the two unit vectors. then there's an orthogonal great circle passing through their midpoint

and then one can pick a unique great circle is perpendicular to both

the latter two great circles are the two cases of equality

also, my choice of $\hat{a},\hat{b}$ was very much deliberate to make the formulas nice. i know they won't be in general

that said, i think i can boil it down in a coordinate-independent way: $$(\hat{n}\cdot \hat{a})^2 -(\hat{n}\cdot \hat{b})^2 =\frac14 (\hat{n}\cdot(\hat{a}+\hat{b}))(\hat{n}\cdot(\hat{a}-\hat{b}))$$

so it's a matter of whether both components in the $\hat{a}+\hat{b},\hat{a}-\hat{b}$ directions have the same sign

oh. that 1/4 is spurious and should be ignored

okay, i think i get why i was confused

so the above should be correct

Let $(M,g)$ be a submanifold of $(\widetilde{M}, \widetilde{g})$ with LC connections $\nabla$ and $\widetilde{\nabla}$. Choose a normal field $\nu$ on $M$. Define the second fundamental form of $M$ as $II(X, Y) = (\widetilde{\nabla}_X Y)^T = \widetilde{\nabla}_X Y - \nabla_X Y$, and the scalar second fundamental form as $h(X, Y) = \langle II(X, Y), \nu \rangle$.

where by $\langle , \rangle$, I mean the metric.

I am trying to show the scalar second fundamental form is equal to $\langle \widetilde{\nabla}_X \nu, Y \rangle$.

This is saying that taking the covariant derivative of $\nu$ along $X$ and then projecting it along $Y$ is the same as taking the covariant derivative of $Y$ and the projecting it along $\nu$.

I guess this should follow from the metric compatibility of $\widetilde{\nabla}$ but I am not sure how.