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Texts on GA seem to start out saying "we're going to replace vector analysis with something intuitive and natural and eloquent and..." and then immediately intr

by ajkjk 8y ago

Texts on GA seem to start out saying "we're going to replace vector analysis with something intuitive and natural and eloquent and..." and then immediately introduce the "geometric product", which is neither intuitive, eloquent, or, as far as I can tell, natural. Its properties are found via formula-wrangling, and the results you get from it seem to just magically work, rather than being intuitive. Bivectors in general are conceptually great; it's just the geometric product which I think is conceptually flimsy. And it's surprisingly hard to compute or make sense of the geometric product in general which is why the writer of the first book cited on the OP wrote "I do not think it possible to give a quick definition of the general geometric product."[1] As far as I can tell 95% of the usefulness of geometric algebra is the usefulness of the wedge product ∧, which is absolutely under-appreciated and appears in loads of places in disguise (for instance, the determinant of a matrix is the wedge product of all of its rows or columns together). The last 5% of the usefulness of GA comes from using the geometric product in vector-rotation via -ava^-1, which is admittedly very useful (it's why physics and computer graphics represent rotation this way, albeit in disguised forms like Pauli matrices and quaternions.) You can express that form without the geometric product, but I haven't found a way that strikes me as very elegant. I don't mean to condemn GA - I think what it's doing is massively important. The mathematical language for vector analysis deficient compared to what we could be using, and a lot of things are more intuitive and natural in better language. I just suspect that the, uh, ideal form of this stuff will look slightly different than GA, and might not include the geometric product at all, but will definitely include the wedge product absolutely everywhere. (I've spent a lot of free time trying to figure this out but I don't really have a compelling result yet. I've been meaning to try blogging about it, though, since it's basically my favorite thing to study.) [1] https://math.stackexchange.com/questions/444988/looking-for-a-clear-definition-of-the-geometric-product https://math.stackexchange.com/questions/444988/looking-for-...

marctenbosch 8y ago
Yeah, I spent a long time trying to find a very clean path to the geometric product and failed so far, but I feel I am getting closer, ahah.
tomjakubowski 8y ago
It's very cool that the geometric product of any orthonormal basis for R^2 gives i as a "pseudoscalar". Let {e1,e2} be an orthonormal basis for R^2. Let i = e1e2. Then i^2 = e1e2e1e2 = -e1e1e2e2 (anticommutivity) = -(1)(1) = -1. Personally that alone is enough to justify learning more about geometric algebra.
- ajkjk 8y ago
  That is true of the wedge product also, though. That's my point. The geometric product gives very little power over the wedge product, and when it does, it's massively non-intuitive how it works.
  - jacobolus 8y ago
    The wedge product of a bivector with itself is 0. To get your bivector to square to –1, you need to be using the geometric product.
  - DoctorOetker 8y ago
    One reason the geometric product is attractive over usual vector notation is for the same reason the usual vector notation is attractive over coordinate notations (think Maxwell Equations in terms of Ex, Ey, Ez etc [0]): instead of a verbose and highly repetitive system of equations with easy operators you can reason and calculate with a single equation, but where the operators have higher complexity, but also correspondingly more properties to exploit assuming you take the time to get familiarized with them. An example and another reason is that contary to the inner product and exterior product, the geometric product is associative, which is clearly useful! The student who is being advised to learn integrals might similarily protest that "everything people love about integrals, could much more easily be understood by just the desirable parts: limits, of summations, of products. It is unnecessary overkill to learn the theory of integrals like int(f(x), x=a..b)+int(f(x),x=b..c)=int(f(x),x=a..c)". So yes to prove the theory of integrals you will need to understand limits, sums, products, .. but the resulting properties like the identity above are undeniably invaluable ... Imagine being a student in an alternate history, where integrals were never defined, of course they can still derive all the results (minus the results stating things about integrals themselves) which we arrive at through our current application of integrals by means of limits of sums of products. But then every derivation that in our world would sanely use integrals would be a long verbose derivation, which they might call "limited totalizations" without reifying this concept. So they are basically rederiving the same result over and over. That's when people normally add syntactic sugar to avoid repitition. Now imagine being this student following such a course, and further imagine that before following this course you had always been somewhat of an autodidact in high school etc, so half the time you are reading your course notes and half the time you are reading books out of curriculum. One day you stumble on some "integer math" book, and then while reading you realize this is not about integers, so you reread the title and it actually reads "math of integrals". After reading on you realize that all the verbose and highly redundant notation in "limited totalization" class can be avoided. Thats what I experienced: I was reading random books about "algebraic geometry" and one of them was totally whacky and off and not algebraic geometry, then I notice the title is actually "geometric algebra". I was in my 3rd year physics. So the people who are paid to teach me are giving me shitty calculitis, most of them are simply unaware of this field, some of them are but are daunted or simply lack the time to go back through all the knowledge they have learnt and rephrase them into this language, and even if they could, it would require the whole curriculum to change in "sync" (well, with a delay of one year per generation...). It really is inertia. The number of people who have come to understand and use geometric algebra are simply fewer than the number who have come to understand and use normal linear algebra, hence there are more books on normal linear algebra. Just like the number of people who have learnt to read and write is larger than the number of people who understand linear algebra, and hence there are more fiction books, magazines... However I believe the situation is slowly changing in the right direction, computer science didn't have compilers either for a while, sooner or later people get bored of spaghetti code... [0] A funny anecdote is that Maxwell -the king of unification in physics- was forced by his publisher to dumb down to this coordinate notation. His original submission used Hamilton quaternions, which had also been recognized by Clifford, who had based his work off of Grassman's "geometric algebra". Clifford called the algebra "geometric algebra", but readers of Clifford started calling the subject of Clifford's work "Clifford algebra". Or something like that, I don't pedantically check history claims...
vanderZwan 8y ago
> As far as I can tell 95% of the usefulness of geometric algebra is the usefulness of the wedge product ∧, which is absolutely under-appreciated and appears in loads of places in disguise (for instance, the determinant of a matrix is the wedge product of all of its rows or columns together). Also, it's really intuitive. At least I found it really intuitive when it was first explained to me.
eigenspace 8y ago
From the sounds of it, I think you'd be much more happy working with differential forms! Differential forms use only the wedge product and a duality mapping known as the hodge dual (It's the GA equivalent of multiplying by the psuedoscalar). Differnetial forms easily supply 90% of the nice things in GA, so if you're unhappy with the geomtric product, I'd say that differnetial forms are perfect. Geometric algebra will be here waiting for when you get tired of resorting to the hodge dual to take an inner product ;)
- ajkjk 8y ago
  I learned about differential forms in my physics degree, and then later discovered that the exterior algebra made a lot more sense if you introduced it separately. Basically it seems like the exterior algebra is useful in many settings, and differential-forms are just one rather confusing application of it. My current opinion is that exterior algebra should, someday, be taught first (around the time of linear algebra), and then later extended to differential forms when derivatives and manifolds get involved. I agree that the Hodge Dual is easily the worst part of exterior algebra. But you can treat the inner product as more fundamental, via *a ∧ b = <a,b> i. Either can essentially be constructed from the other (iirc).
  - cygx 8y ago
    The Hodge star isn't so bad: It's just the generalization of the idea that you can describe a plane by its normal vector.
    - ajkjk 8y ago
      On it's own it's not bad, but it's troublesome because given a complicated expression with wedges and stars, it's not at all intuitive how you can simplify it without working in coordinates.
      - cygx 8y ago
        Is there a silver bullet, though? The complexity of having to mess around with epsilon tensors is still there no matter which abstraction you use to sweep it under the rug...
yiyus 8y ago
In my opinion, the problem is not the geometric product, but having three products (inner, outer and geometric). I have explained GA using a single product, and I think it is much clearer. Of course, the inner and outer products are also very useful, but I introduce them only at the last stage, as some extra notation, not to help understand new concepts.
jacobolus 8y ago
The geometric product is super useful and important! (As well as much nicer to work with when doing algebraic manipulation.) In particular, it is what lets you take products, inverses and quotients (assuming the denominator is non-null) of arbitrary vectors. > surprisingly hard to compute Hm? No it isn’t.... Several times in the last few years I have done several pages of complicated calculations in terms of coordinates or matrices, where the intermediate steps were basically an indecipherable mess of symbols, and then sat down, really thought about what I was doing for a while, and re-written it using GA language and ended up simplifying my work down to a few lines of simple algebraic manipulations, where I could typically then give a nice geometrical interpretation to each step. If you want to compute the geometric product using concrete numbers in terms of a basis, then it is straightforward to write the code (or work through by hand). What is hard is not the geometric product. What is hard is geometry! There is a lot of depth and subtlety, and it takes years to really learn your way around. That is not the fault of the language though; there is a certain amount of irreducible complexity involved. Personally I would love to get some people together and try to write a high-school-accessible geometry / vector algebra book using GA language. It would be a few years of hard work, because figuring out the right order for the curriculum, the right types of problems to build intuition, which tools to include or leave out of an introductory book, how to best show the work for a whole collection of worked examples (ideally with some interactive computer diagrams), and so on.
- ajkjk 8y ago
  It is easy to compute on vectors. It is possible, but complicated and unintuitive (I have written code to do it in the past), to compute it on multivectors. I'm arguing in favor of most of GA's language. I just keep finding that the wedge and inner product parts are fantastic, and the geometric product part isn't. And I think the reason people keep finding GA appealing is because they didn't have the wedge product before, so having that in their conceptual toolkit fixes a lot, while having the geometric product doesn't fix much on top of it. Anyway I've studied GA a lot and I still have basically no idea what 'AB' means when both are arbitrary-grade multivectors, and as far as I can tell most sources don't even try to explain it.
  - 8y ago
ecl3ctic 8y ago
Please do blog about this. I'd be really interested to hear your thoughts!
jacobolus 8y ago
By the way, the determinant of a matrix is just the outermorphism of the associated linear transformation applied to a unit pseudoscalar. The outermorphism of a linear transformation is a useful and convenient (but sometimes tricky) concept. It gives you not only determinants, but also the application of your linear transformation to arbitrary multivectors.
- IshKebab 8y ago
  No the determinant is the isomorphism of the third latent moment of the orthonorm. Duh.
  - jacobolus 8y ago
    https://en.wikipedia.org/wiki/Outermorphism https://en.wikipedia.org/wiki/Outermorphism
infinity0 8y ago
The way this text does the geometric product is trying to be intuitive by immediately giving a "physical" interpretation as the sum of the inner and outer products. However the physical interpretations breaks down when trying to multiply three of these together. I think Ben Lynn's approach is actually better: https://crypto.stanford.edu/~blynn/haskell/ga.html https://crypto.stanford.edu/~blynn/haskell/ga.html (Part 1) It starts off more abstract, defining the geometric product simply as string concatenation (as in "free monoid" if you're familiar with that term, which you would if you have intermediate Haskell knowledge) plus a very natural constraint. From this natural constraint one can then deduce the sum-based interpretation that the OP gives. The natural constraint is then further generalised and justified in more detail in Part 3: https://crypto.stanford.edu/~blynn/haskell/cga.html https://crypto.stanford.edu/~blynn/haskell/cga.html

by ajkjk 8y ago

marctenbosch 8y ago

Yeah, I spent a long time trying to find a very clean path to the geometric product and failed so far, but I feel I am getting closer, ahah.

tomjakubowski 8y ago

It's very cool that the geometric product of any orthonormal basis for R^2 gives i as a "pseudoscalar". Let {e1,e2} be an orthonormal basis for R^2. Let i = e1e2. Then i^2 = e1e2e1e2 = -e1e1e2e2 (anticommutivity) = -(1)(1) = -1. Personally that alone is enough to justify learning more about geometric algebra.

ajkjk 8y ago
That is true of the wedge product also, though. That's my point. The geometric product gives very little power over the wedge product, and when it does, it's massively non-intuitive how it works.
- jacobolus 8y ago
  The wedge product of a bivector with itself is 0. To get your bivector to square to –1, you need to be using the geometric product.
- DoctorOetker 8y ago
  One reason the geometric product is attractive over usual vector notation is for the same reason the usual vector notation is attractive over coordinate notations (think Maxwell Equations in terms of Ex, Ey, Ez etc [0]): instead of a verbose and highly repetitive system of equations with easy operators you can reason and calculate with a single equation, but where the operators have higher complexity, but also correspondingly more properties to exploit assuming you take the time to get familiarized with them. An example and another reason is that contary to the inner product and exterior product, the geometric product is associative, which is clearly useful! The student who is being advised to learn integrals might similarily protest that "everything people love about integrals, could much more easily be understood by just the desirable parts: limits, of summations, of products. It is unnecessary overkill to learn the theory of integrals like int(f(x), x=a..b)+int(f(x),x=b..c)=int(f(x),x=a..c)". So yes to prove the theory of integrals you will need to understand limits, sums, products, .. but the resulting properties like the identity above are undeniably invaluable ... Imagine being a student in an alternate history, where integrals were never defined, of course they can still derive all the results (minus the results stating things about integrals themselves) which we arrive at through our current application of integrals by means of limits of sums of products. But then every derivation that in our world would sanely use integrals would be a long verbose derivation, which they might call "limited totalizations" without reifying this concept. So they are basically rederiving the same result over and over. That's when people normally add syntactic sugar to avoid repitition. Now imagine being this student following such a course, and further imagine that before following this course you had always been somewhat of an autodidact in high school etc, so half the time you are reading your course notes and half the time you are reading books out of curriculum. One day you stumble on some "integer math" book, and then while reading you realize this is not about integers, so you reread the title and it actually reads "math of integrals". After reading on you realize that all the verbose and highly redundant notation in "limited totalization" class can be avoided. Thats what I experienced: I was reading random books about "algebraic geometry" and one of them was totally whacky and off and not algebraic geometry, then I notice the title is actually "geometric algebra". I was in my 3rd year physics. So the people who are paid to teach me are giving me shitty calculitis, most of them are simply unaware of this field, some of them are but are daunted or simply lack the time to go back through all the knowledge they have learnt and rephrase them into this language, and even if they could, it would require the whole curriculum to change in "sync" (well, with a delay of one year per generation...). It really is inertia. The number of people who have come to understand and use geometric algebra are simply fewer than the number who have come to understand and use normal linear algebra, hence there are more books on normal linear algebra. Just like the number of people who have learnt to read and write is larger than the number of people who understand linear algebra, and hence there are more fiction books, magazines... However I believe the situation is slowly changing in the right direction, computer science didn't have compilers either for a while, sooner or later people get bored of spaghetti code... [0] A funny anecdote is that Maxwell -the king of unification in physics- was forced by his publisher to dumb down to this coordinate notation. His original submission used Hamilton quaternions, which had also been recognized by Clifford, who had based his work off of Grassman's "geometric algebra". Clifford called the algebra "geometric algebra", but readers of Clifford started calling the subject of Clifford's work "Clifford algebra". Or something like that, I don't pedantically check history claims...

vanderZwan 8y ago

> As far as I can tell 95% of the usefulness of geometric algebra is the usefulness of the wedge product ∧, which is absolutely under-appreciated and appears in loads of places in disguise (for instance, the determinant of a matrix is the wedge product of all of its rows or columns together). Also, it's really intuitive. At least I found it really intuitive when it was first explained to me.

eigenspace 8y ago

From the sounds of it, I think you'd be much more happy working with differential forms! Differential forms use only the wedge product and a duality mapping known as the hodge dual (It's the GA equivalent of multiplying by the psuedoscalar). Differnetial forms easily supply 90% of the nice things in GA, so if you're unhappy with the geomtric product, I'd say that differnetial forms are perfect. Geometric algebra will be here waiting for when you get tired of resorting to the hodge dual to take an inner product ;)

ajkjk 8y ago
I learned about differential forms in my physics degree, and then later discovered that the exterior algebra made a lot more sense if you introduced it separately. Basically it seems like the exterior algebra is useful in many settings, and differential-forms are just one rather confusing application of it. My current opinion is that exterior algebra should, someday, be taught first (around the time of linear algebra), and then later extended to differential forms when derivatives and manifolds get involved. I agree that the Hodge Dual is easily the worst part of exterior algebra. But you can treat the inner product as more fundamental, via *a ∧ b = <a,b> i. Either can essentially be constructed from the other (iirc).
- cygx 8y ago
  The Hodge star isn't so bad: It's just the generalization of the idea that you can describe a plane by its normal vector.
  - ajkjk 8y ago
    On it's own it's not bad, but it's troublesome because given a complicated expression with wedges and stars, it's not at all intuitive how you can simplify it without working in coordinates.
    - cygx 8y ago
      Is there a silver bullet, though? The complexity of having to mess around with epsilon tensors is still there no matter which abstraction you use to sweep it under the rug...

yiyus 8y ago

In my opinion, the problem is not the geometric product, but having three products (inner, outer and geometric). I have explained GA using a single product, and I think it is much clearer. Of course, the inner and outer products are also very useful, but I introduce them only at the last stage, as some extra notation, not to help understand new concepts.

jacobolus 8y ago

The geometric product is super useful and important! (As well as much nicer to work with when doing algebraic manipulation.) In particular, it is what lets you take products, inverses and quotients (assuming the denominator is non-null) of arbitrary vectors. > surprisingly hard to compute Hm? No it isn’t.... Several times in the last few years I have done several pages of complicated calculations in terms of coordinates or matrices, where the intermediate steps were basically an indecipherable mess of symbols, and then sat down, really thought about what I was doing for a while, and re-written it using GA language and ended up simplifying my work down to a few lines of simple algebraic manipulations, where I could typically then give a nice geometrical interpretation to each step. If you want to compute the geometric product using concrete numbers in terms of a basis, then it is straightforward to write the code (or work through by hand). What is hard is not the geometric product. What is hard is geometry! There is a lot of depth and subtlety, and it takes years to really learn your way around. That is not the fault of the language though; there is a certain amount of irreducible complexity involved. Personally I would love to get some people together and try to write a high-school-accessible geometry / vector algebra book using GA language. It would be a few years of hard work, because figuring out the right order for the curriculum, the right types of problems to build intuition, which tools to include or leave out of an introductory book, how to best show the work for a whole collection of worked examples (ideally with some interactive computer diagrams), and so on.

ajkjk 8y ago
It is easy to compute on vectors. It is possible, but complicated and unintuitive (I have written code to do it in the past), to compute it on multivectors. I'm arguing in favor of most of GA's language. I just keep finding that the wedge and inner product parts are fantastic, and the geometric product part isn't. And I think the reason people keep finding GA appealing is because they didn't have the wedge product before, so having that in their conceptual toolkit fixes a lot, while having the geometric product doesn't fix much on top of it. Anyway I've studied GA a lot and I still have basically no idea what 'AB' means when both are arbitrary-grade multivectors, and as far as I can tell most sources don't even try to explain it.
- 8y ago

ecl3ctic 8y ago

Please do blog about this. I'd be really interested to hear your thoughts!

jacobolus 8y ago

By the way, the determinant of a matrix is just the outermorphism of the associated linear transformation applied to a unit pseudoscalar. The outermorphism of a linear transformation is a useful and convenient (but sometimes tricky) concept. It gives you not only determinants, but also the application of your linear transformation to arbitrary multivectors.

IshKebab 8y ago
No the determinant is the isomorphism of the third latent moment of the orthonorm. Duh.
- jacobolus 8y ago
  https://en.wikipedia.org/wiki/Outermorphism https://en.wikipedia.org/wiki/Outermorphism

infinity0 8y ago

The way this text does the geometric product is trying to be intuitive by immediately giving a "physical" interpretation as the sum of the inner and outer products. However the physical interpretations breaks down when trying to multiply three of these together. I think Ben Lynn's approach is actually better: https://crypto.stanford.edu/~blynn/haskell/ga.html https://crypto.stanford.edu/~blynn/haskell/ga.html (Part 1) It starts off more abstract, defining the geometric product simply as string concatenation (as in "free monoid" if you're familiar with that term, which you would if you have intermediate Haskell knowledge) plus a very natural constraint. From this natural constraint one can then deduce the sum-based interpretation that the OP gives. The natural constraint is then further generalised and justified in more detail in Part 3: https://crypto.stanford.edu/~blynn/haskell/cga.html https://crypto.stanford.edu/~blynn/haskell/cga.html