“what is the nature of the connection between orthonormality and squaring”
I have a different answer to go alongside AlexMennen’s answer.
In differential topology, there is an important distinction between vectors and covectors. To see what this is, we need to look at the behavior under a change of basis. If we double our basis vectors, then we’ll need to halve the coordinates of a vector, but we’ll need to double the coordinates of a covector. A good way to visualize this is as a geographical map with contour lines. Position differences are vectors, and the contours are covectors.
You can think of covectors as measuring vectors, but without adding something new, there’s not a natural way to compare two vectors to each other. To compare two things, you need a function that will take those two things as an input, and return a scalar. In order for such a function to be invariant under change of basis, it will have to be a (0, 2)-tensor (aka a bilinear form). Let’s call this tensor T(u,v). Now if we multiply u and v by a scalar r, then bilinearity forces that T(ru, rv) = r^2 T(u,v) which is the squaring we were looking form (and in general, you can prove that T(u,u) is a quadratic form, and that it must actually be quadratic if we want both inputs to matter).
So to summarize:
Orthogonality is inherently a way to compare two things of the same type.
To compare 2 things of the same type in a differential space inherently requires a bilinear form.
Non-trivial bilinear forms are inherently quadratic.
Thanks! This seems to me like another piece of the puzzle =D
In this case, this is one that I already had (at least, well enough for the hindsight bias to kick in :-p), and it’s on my list of trailheads next time I try to ground out the 2 in the Born rule. FWIW, some lingering questions I have when I take this viewpoint include “ok, cool, why are there no corresponding situations where I want to compare 3 vectorish thingies?” / “I see why the argument works for 2, but I have a sneaking suspicion that this 2 is being slipped into the problem statement in a way that I’m not yet quite following”. Also, I have a sense that there’s some fairly important fact about anointing some linear isomorphism between a vector space and its dual as “canonical” that I have yet to grasp. Like, some part of the answer to “why do I never want to compare 3 vectorish thingies” is b/c the relationship between a vector space and its dual space is somehow pretty special, and there’s no correspondingly special… triality of vector spaces. (Hrm. I wonder whether I can ground the 2 in the Born rule out into the 2 in categorical duality. That would be nuts.)
FWIW, one of my litmus tests is question “assuming we are supposed to measure distance in R2 using an L2 norm, why are we not supposed to measure distance in R3 using an L3 norm?”. And, like, I have a bunch of explanations for why this is (including “L3 isn’t invariant under most any change of basis” (per Alex’s argument above) and “b/c the natural notion of distance between two points in R3 factors into two questions of distance in R2, so using L2 in R2 pins down using L2 in Rn”), but I still feel like there’s some… surprising fixation on 2 here, that I can’t yet explain to the satisfaction of young-Nate who guesses that cuberute(x^3 + y^3 + z^3)=r is the equation for a sphere. Like, I still feel kinda like math said “proof by induction: starting in the case where n=2, …” and I’m like “wat why aren’t we starting at 0” and it’s like “don’t worry, n < 2 will work out as special cases” and I’m like “ok, sure, this argument is valid, but also wtf are you doing”. My wtfs aren’t all ironed out yet.
And, maybe I’m chasing shadows (eg, seeking a logical explanation where there is none, which is the sort of thing that can happen to a poor sap who lacks an explicit understanding of logical causality), but my suspicion is that I’m still missing part of the explanation. And both this route (which I’d gloss as “understand why it’s so important/natural/??? to have/choose/determine a canonical isomorphism between your vector space and its dual, then appeal to bilinearity”, with a prereq of better understanding why we have/care-about duality but not triality in Vect) and Alex’s (which I’d gloss as “explain why 2 obviously-should-be the balance point in the Lp norms”) both feel like good trailheads to me.
(And, in case this wasn’t clear to all readers, none of my assertions of confusion here are allegations that everyone else is similarly confused—indeed, Alex and Adele have already given demonstrations to the contrary.)
So, trilinear forms are a thing: for example, if you have 3 vectors, and you want to know the volume of the parallelepiped they form, that’s a trilinear form. And that clearly has a “cubicness” to it, and you can do this for arbitrary numbers of vectors and covectors. The Riemann curvature tensor is perhaps the most significant one that has more than 2 (co)vectors involved. FWIW the dual space thing also seems likely to be important for my confusion about why phase space “volume” is 2-dimensional (even in super huge phase spaces)!
I would say that distance is bilinear in arbitrary dimension because it’s also inherently a comparison of two vectors (a vector to measure, and a “unit” vector to measure it by). Not sure if that reduces things any for you.
For me, it doesn’t feel like there’s going to be anything beyond “because comparison is important, and inherently 2-ish” for this. I do think part of why a metric is so significant is related to the dual space, but my guess is that even this will ultimately boil down to “comparison” (maybe as the concept of equality) being important.
I have a different answer to go alongside AlexMennen’s answer.
In differential topology, there is an important distinction between vectors and covectors. To see what this is, we need to look at the behavior under a change of basis. If we double our basis vectors, then we’ll need to halve the coordinates of a vector, but we’ll need to double the coordinates of a covector. A good way to visualize this is as a geographical map with contour lines. Position differences are vectors, and the contours are covectors.
You can think of covectors as measuring vectors, but without adding something new, there’s not a natural way to compare two vectors to each other. To compare two things, you need a function that will take those two things as an input, and return a scalar. In order for such a function to be invariant under change of basis, it will have to be a (0, 2)-tensor (aka a bilinear form). Let’s call this tensor T(u,v). Now if we multiply u and v by a scalar r, then bilinearity forces that T(ru, rv) = r^2 T(u,v) which is the squaring we were looking form (and in general, you can prove that T(u,u) is a quadratic form, and that it must actually be quadratic if we want both inputs to matter).
So to summarize:
Orthogonality is inherently a way to compare two things of the same type.
To compare 2 things of the same type in a differential space inherently requires a bilinear form.
Non-trivial bilinear forms are inherently quadratic.
Thanks! This seems to me like another piece of the puzzle =D
In this case, this is one that I already had (at least, well enough for the hindsight bias to kick in :-p), and it’s on my list of trailheads next time I try to ground out the 2 in the Born rule. FWIW, some lingering questions I have when I take this viewpoint include “ok, cool, why are there no corresponding situations where I want to compare 3 vectorish thingies?” / “I see why the argument works for 2, but I have a sneaking suspicion that this 2 is being slipped into the problem statement in a way that I’m not yet quite following”. Also, I have a sense that there’s some fairly important fact about anointing some linear isomorphism between a vector space and its dual as “canonical” that I have yet to grasp. Like, some part of the answer to “why do I never want to compare 3 vectorish thingies” is b/c the relationship between a vector space and its dual space is somehow pretty special, and there’s no correspondingly special… triality of vector spaces. (Hrm. I wonder whether I can ground the 2 in the Born rule out into the 2 in categorical duality. That would be nuts.)
FWIW, one of my litmus tests is question “assuming we are supposed to measure distance in R2 using an L2 norm, why are we not supposed to measure distance in R3 using an L3 norm?”. And, like, I have a bunch of explanations for why this is (including “L3 isn’t invariant under most any change of basis” (per Alex’s argument above) and “b/c the natural notion of distance between two points in R3 factors into two questions of distance in R2, so using L2 in R2 pins down using L2 in Rn”), but I still feel like there’s some… surprising fixation on 2 here, that I can’t yet explain to the satisfaction of young-Nate who guesses that cuberute(x^3 + y^3 + z^3)=r is the equation for a sphere. Like, I still feel kinda like math said “proof by induction: starting in the case where n=2, …” and I’m like “wat why aren’t we starting at 0” and it’s like “don’t worry, n < 2 will work out as special cases” and I’m like “ok, sure, this argument is valid, but also wtf are you doing”. My wtfs aren’t all ironed out yet.
And, maybe I’m chasing shadows (eg, seeking a logical explanation where there is none, which is the sort of thing that can happen to a poor sap who lacks an explicit understanding of logical causality), but my suspicion is that I’m still missing part of the explanation. And both this route (which I’d gloss as “understand why it’s so important/natural/??? to have/choose/determine a canonical isomorphism between your vector space and its dual, then appeal to bilinearity”, with a prereq of better understanding why we have/care-about duality but not triality in Vect) and Alex’s (which I’d gloss as “explain why 2 obviously-should-be the balance point in the Lp norms”) both feel like good trailheads to me.
(And, in case this wasn’t clear to all readers, none of my assertions of confusion here are allegations that everyone else is similarly confused—indeed, Alex and Adele have already given demonstrations to the contrary.)
<3 hooray.
Awesome!
So, trilinear forms are a thing: for example, if you have 3 vectors, and you want to know the volume of the parallelepiped they form, that’s a trilinear form. And that clearly has a “cubicness” to it, and you can do this for arbitrary numbers of vectors and covectors. The Riemann curvature tensor is perhaps the most significant one that has more than 2 (co)vectors involved. FWIW the dual space thing also seems likely to be important for my confusion about why phase space “volume” is 2-dimensional (even in super huge phase spaces)!
I would say that distance is bilinear in arbitrary dimension because it’s also inherently a comparison of two vectors (a vector to measure, and a “unit” vector to measure it by). Not sure if that reduces things any for you.
For me, it doesn’t feel like there’s going to be anything beyond “because comparison is important, and inherently 2-ish” for this. I do think part of why a metric is so significant is related to the dual space, but my guess is that even this will ultimately boil down to “comparison” (maybe as the concept of equality) being important.