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Ah, another geometric algebra evangelist? I can't figure out if GA actually adds anything substantial, or if it merely lets us write some equations in a more su

Ah, another geometric algebra evangelist? I can't figure out if GA actually adds anything substantial, or if it merely lets us write some equations in a more succinct fashion. But it certainly looks cool. As to vectors, obviously they have inverses, additive inverses. Since vectors don't have multiplication, there is no multiplicative inverse, but if you define new operations on them, well then that operation can have an inverse but that is not really "the inverse of a vector" anymore.

thaumasiotes 5y ago
> As vectors, obviously they have inverses, additive inverses. Since vectors don't have multiplication, there is no multiplicative inverse A vector is pretty much by definition also a matrix, and there is a standard way to multiply matrices. You can define several inverses of a vector that way, though you can't define a unique inverse. The standard inner product is of course also an exceptionally typical way to multiply vectors, but the concept of an inverse there doesn't make much sense.
- monktastic1 5y ago
  > The standard inner product is of course also an exceptionally typical way to multiply vectors, but the concept of an inverse there doesn't make much sense. Not all vector spaces are equipped with an inner product. The point is that you can start with some simple axioms and build these more complicated things (inner product spaces, algebras over a field, geometric algebras, etc.).
  - thaumasiotes 5y ago
    > Not all vector spaces are equipped with an inner product. Any vector space over a field (usually part of the definition of a vector space) is equipped with the standard inner product, because multiplication and addition are part of the definition of a field.
    - Grustaf 5y ago
      That’s not how it works. Unless you explicitly give the vector space an inner product, it doesn’t have one. What you probably mean is that “you can always define an inner product” but that’s a very different statement. That’s not true either though, the different dimensions in a vector space do bot have to belong to the same field, so you can’t assume you can add them together.
      - thaumasiotes 5y ago
        > Unless you explicitly give the vector space an inner product, it doesn’t have one. Well, no, not at all. The inner product is still there. It's still an inner product. The space in which your vectors exist is still an inner product space. You may not care about the inner product, but it doesn't cease to exist when you stop looking at it.
        Grustaf 5y ago
        I really don't understand why you would say this, it's obviously false. Setting aside the subtler point of what it means to "have" something in mathematics: clearly only some vector spaces even have the potential to introduce an inner product. Consider F for some random finite field. You can make a vector space from it, but what would the inner product be? Or R x F for that matter, you could never give that an inner product. That's why the concepts of "vector space" and "inner product space" are separate concepts. Some vector spaces aren't, and could never be, inner product spaces.
        monktastic1 5y ago
        https://math.stackexchange.com/a/247438/161555 https://math.stackexchange.com/a/247438/161555
- Grustaf 5y ago
  No, a vector is defined as an object that has certain properties, like addition and scalar multiplication. It's a very general, and abstract concept. There are vector spaces of functions, with infinite dimension, but there are also vector spaces with a finite number of elements. So only some vectors can even be written as 1xN matrices, if that is what you're referring to. But even if you write a vector that way, it doesn't mean it IS a matrix or that it automatically "has" multiplication. In mathematics, an object only has an operation if it's part of the definition, and as such, vectors don't "have" multiplication.
  - adrian_b 5y ago
    What you say is mostly correct, but only for a certain meaning of the word "vector", which has been used with 2 distinct meanings since its introduction in the first half of the 19th century. The set of elements defined by certain properties of their addition and of their multiplication with the elements belonging to a set of scalars is named "vector space" by some and "linear space" by others. According to the etymology of the word vector, "linear space" would be more appropriate. You have used "vector" with the meaning "element of a linear space", and what you have said is correct, except that for any "vector" as an element of a linear space, considered as a column vector, there exists a corresponding row vector, even in the infinite-dimensional case. "Vector" means translation of the space, and this is what "vector" meant when the word was introduced by Hamilton. While the set of translations is a linear space a.k.a. a vector space in the generalized sense, the set of translations, i.e. vectors in the strict sense, has additional properties due to the multiplication operations that must be defined for "vectors" in their strict sense (which are needed e.g. to determine the angles between translations and the distances). "Vectors" as elements of linear spaces are a very general notion, which appears in many domains, and for all linear spaces, including for those infinite-dimensional, you can define matrices, i.e. linear functions, and matrix multiplication, i.e. composition of linear functions, and also the correspondence between a 1xN vector and a Nx1 vector, more correctly between a vector and an associated linear form. The latter also exists for the infinite-dimensional case, even if it is less likely to use names like row vectors and column vectors (though the names bra vectors and ket vectors are still in use for the infinite-dimensional case). For the infinite-dimensional case the vectors and the matrices become functions of 1 or of 2 parameters and the sums from the formulas of matrix multiplication become integrals. While for most computer applications, "vectors" refer just to elements of linear spaces, most "vectors" used in models of physical systems are vectors in the original sense of the word, where not only the vector addition and the product with scalars matter, but the products of vectors also have an essential role and their meaning can be best understood in the context of the complete geometric algebra theory.
- ReleaseCandidat 5y ago
  > A vector is pretty much by definition also a matrix, and there is a standard way to multiply matrices. A standard way to multiply a MxN with a NxK matrix, but none for a 1xN with a 1xN or a Nx1 with a Nx1 matrix - the two possible ways to describe a vector. You have to transpose exactly one of the two vectors. And then you have two possible results, 1xN multiplied with Nx1 yields a scalar (that's actually the 'usual' dot-product/scalar product/whatever you call it) and Nx1 multiplied with 1xN, where the result is a NxN matrix.
  - thaumasiotes 5y ago
    Doesn't matter. We already didn't have a unique inverse, but it's perfectly possible to find a left pseudoinverse and a right pseudoinverse, bearing in mind that they're not unique. Though thinking about it more, it seems like the outer-product-inverse of a vector (a) must be unique if it exists; and (b) is highly unlikely to exist. > 1xN multiplied with Nx1 yields a scalar (that's actually the 'usual' dot-product/scalar product/whatever you call it) I'm aware of this, but there are two ways we might conceive of an "inverse": - Since a vector is a matrix, the inverse of a vector might be defined by matrix multiplication, where A is the inverse of B if AB is "the" identity matrix. This is only strictly defined for square matrices, but the pseudoinverse concept extends it to nonsquare matrices. - Or, we could go for a more basic sense of "multiplicative inverse", where the concept is that if AB = C, then B = A⁻¹C. This is what I was thinking of when saying that the concept of an inverse doesn't make sense when multiplication is the inner product - if I give you a vector v, and its inner product with some other vector u, there is no way of recovering what u was.
BlackFly 5y ago
If you are working in ordinary Euclidean space with orthonormal bases, it doesn't do much for you. When you start doing calculus on embedded surfaces (the beginning of differential manifolds) it begins to be more helpful. Since Lie algebras are special types of differential manifolds, you can learn quite a bit by studying the geometry and this quickly leads into gauge theory and modern physics. The Geometry of Physics: An Introduction, by Theodore Frankel does a good job illustrating a lot of aspects of geometry even offering some geometric insight into some classical physics. It is an absolute necessity for general relativity.
- contravariant 5y ago
  Is there some canonical metric on Lie algebras that makes this possible? Otherwise geometric algebra won't get you far.
  - enkimute 5y ago
    Every Lie algebra is a bivector algebra (see 'Lie groups as Spin groups' http://geocalc.clas.asu.edu/pdf/LGasSG.pdf http://geocalc.clas.asu.edu/pdf/LGasSG.pdf). Additionally the GA formalism enables closed form solutions for the exponential map for all bivector algebras. (see 'Graded symmetry Groups' https://www.researchgate.net/publication/353116859_Graded_Symmetry_Groups_Plane_and_Simple/ https://www.researchgate.net/publication/353116859_Graded_Sy...).
- Grustaf 5y ago
  That books seems very interesting, but is it really about Geometric Algebra? I'm not talking about geometry in general, or differential geometry, which I'm a bit familiar with. And not Algebraic Geometry either, for that matter, which is also a fascinating subject. I mean specifically Geometric Algebra. It sort of seems like a notation, but it has almost a cult like following and perhaps it's more than a notation, is it a theory, a branch of mathematics?
  - yuppiemephisto 5y ago
    It really is useful even at the advanced level. Following it far enough leads to the Atiyah-Singer Index Theorem and Hodge Theory. The advantage over the exterior algebra is that you have that and an interior algebra, which leads to many formulas in differential geometry becoming very natural (like Cartan's magic formula).
rsj_hn 5y ago
It's just some of the more advanced theory you'd get from studying modules repackaged a bit. Basically the extra stuff that is usually skipped in first year linear algebra courses are the symmetric and asymmetric (often called exterior) products. These form algebras, of course. The exterior product, or wedge product, has a natural interpretation in terms of signed areas (or volumes) and from this you get the determinant as a volume form. These are the natural generalizations of dot products (inner products) and wedge products (exterior products). You can take a vector and associate it with a 1 form (asymmetric algebra or exterior algebra), and then multiply two vectors to get a 2 form using the standard wedge product, etc. In dimension 3, the space of 2 forms is dual to the space of 1 forms and so you can "multiply" two vectors to get a third vector. That is all that's going on here. Actually a good multi-variable calculus class will cover most of this stuff as you need some motivation for Jacobian volume forms used to calculate areas and volumes under change of basis, and dot/wedge products are useful for generalizations of the Gauss divergence theorem and the generalized fundamental theorem that says the integral over a function, f, on the n-1 dimensional boundary of a shape is the differential of the integral of the shape. Moreover any class on Riemannian geometry will give you all the linear algebra you need as well. One thing I would caution students with is that by using somewhat non-standard jargon they may not understand how to generalize this stuff to n-dimensions, nor will the connections between, say, determinants and wedge-forms be clear, or dot products and angles be fully understood if only the n=3 cases is emphasized. Only in n=3 can you multiply two vectors to get a vector. But fun fact: in dimension 3k you can multiply two k-forms to get a third k-form (as the space of 2k forms is dual to the space of k forms in n=3k). If you think there is this new thing called "geometric algebra" other than usual tensor products, it may not be obvious how things generalize to n != 3.
- Grustaf 5y ago
  > But no, there is nothing new here beyond marketing. Yes, that's my sense too. Of course cross products, wedge products etc make sense and that's just standard mathematics, but the part that I haven't really seen the point of is to form the algebra where all these forms live side by side. It doesn't seem like a useful "fusing", in the way that say the complex plane is. Of course it's very cool that sub-algebras in 2 or 3-space in GA are isomorphic to the complex plane or even quaternions, but it still feels a bit made up. For a concrete example, one youtuber showed how Maxwells equations simplified to a single equation if you introduce an operator that is a combination of div and curl, and also a new kind of physical entity that combines the electrical and magnetic fields. This is of course cool, but what I want to know is if this new operator makes some physical sense, and if the new multi dimensional field has any physical meaning. If they don't, it just seems like a parlour trick. Not saying they actually don't, but I haven't seen any deeper explanations of it.
nocturnial 5y ago
> Ah, another geometric algebra evangelist? Heh, this made me chuckle and this is reason why I read HN comments. I can't help but to picture someone ringing your doorbell early on an sunday and saying: "Hi! Have you considered inverting vectors?" slams door "Who was it?" "It was just another of those damned geometric algebraists"
- Grustaf 5y ago
  Haha that reminds me, got to put back that "No beggars or geometric albraists" sign!
LotusFunctor 5y ago
For me, the substantial thing geometric algebra gave me so far was a newfound appreciation of the seemingly disparate systems: tensors, differential forms, matrix algebra, and also a newfound appreciation of stuff like determinants, conjugate elements in group theory, lie groups and lie algebras, etc., because it helps clarify the relationship between them, and as another user here said, you can get propelled up into some pretty advanced stuff later on (said user mentions the Atiyah-Singer Index Theorem and Hodge theory, but caveat: I've only recently started tacking a crack at the latter. I will say that, OTOH, it's pretty nice to be able to see something like the wiki on Clifford Analysis and realize its familiar territory from geometric calculus).

thaumasiotes 5y ago

> As vectors, obviously they have inverses, additive inverses. Since vectors don't have multiplication, there is no multiplicative inverse A vector is pretty much by definition also a matrix, and there is a standard way to multiply matrices. You can define several inverses of a vector that way, though you can't define a unique inverse. The standard inner product is of course also an exceptionally typical way to multiply vectors, but the concept of an inverse there doesn't make much sense.

monktastic1 5y ago
> The standard inner product is of course also an exceptionally typical way to multiply vectors, but the concept of an inverse there doesn't make much sense. Not all vector spaces are equipped with an inner product. The point is that you can start with some simple axioms and build these more complicated things (inner product spaces, algebras over a field, geometric algebras, etc.).
- thaumasiotes 5y ago
  > Not all vector spaces are equipped with an inner product. Any vector space over a field (usually part of the definition of a vector space) is equipped with the standard inner product, because multiplication and addition are part of the definition of a field.
  - Grustaf 5y ago
    That’s not how it works. Unless you explicitly give the vector space an inner product, it doesn’t have one. What you probably mean is that “you can always define an inner product” but that’s a very different statement. That’s not true either though, the different dimensions in a vector space do bot have to belong to the same field, so you can’t assume you can add them together.
    - thaumasiotes 5y ago
      > Unless you explicitly give the vector space an inner product, it doesn’t have one. Well, no, not at all. The inner product is still there. It's still an inner product. The space in which your vectors exist is still an inner product space. You may not care about the inner product, but it doesn't cease to exist when you stop looking at it.
      - Grustaf 5y ago
        I really don't understand why you would say this, it's obviously false. Setting aside the subtler point of what it means to "have" something in mathematics: clearly only some vector spaces even have the potential to introduce an inner product. Consider F for some random finite field. You can make a vector space from it, but what would the inner product be? Or R x F for that matter, you could never give that an inner product. That's why the concepts of "vector space" and "inner product space" are separate concepts. Some vector spaces aren't, and could never be, inner product spaces.
      - monktastic1 5y ago
        https://math.stackexchange.com/a/247438/161555 https://math.stackexchange.com/a/247438/161555
Grustaf 5y ago
No, a vector is defined as an object that has certain properties, like addition and scalar multiplication. It's a very general, and abstract concept. There are vector spaces of functions, with infinite dimension, but there are also vector spaces with a finite number of elements. So only some vectors can even be written as 1xN matrices, if that is what you're referring to. But even if you write a vector that way, it doesn't mean it IS a matrix or that it automatically "has" multiplication. In mathematics, an object only has an operation if it's part of the definition, and as such, vectors don't "have" multiplication.
- adrian_b 5y ago
  What you say is mostly correct, but only for a certain meaning of the word "vector", which has been used with 2 distinct meanings since its introduction in the first half of the 19th century. The set of elements defined by certain properties of their addition and of their multiplication with the elements belonging to a set of scalars is named "vector space" by some and "linear space" by others. According to the etymology of the word vector, "linear space" would be more appropriate. You have used "vector" with the meaning "element of a linear space", and what you have said is correct, except that for any "vector" as an element of a linear space, considered as a column vector, there exists a corresponding row vector, even in the infinite-dimensional case. "Vector" means translation of the space, and this is what "vector" meant when the word was introduced by Hamilton. While the set of translations is a linear space a.k.a. a vector space in the generalized sense, the set of translations, i.e. vectors in the strict sense, has additional properties due to the multiplication operations that must be defined for "vectors" in their strict sense (which are needed e.g. to determine the angles between translations and the distances). "Vectors" as elements of linear spaces are a very general notion, which appears in many domains, and for all linear spaces, including for those infinite-dimensional, you can define matrices, i.e. linear functions, and matrix multiplication, i.e. composition of linear functions, and also the correspondence between a 1xN vector and a Nx1 vector, more correctly between a vector and an associated linear form. The latter also exists for the infinite-dimensional case, even if it is less likely to use names like row vectors and column vectors (though the names bra vectors and ket vectors are still in use for the infinite-dimensional case). For the infinite-dimensional case the vectors and the matrices become functions of 1 or of 2 parameters and the sums from the formulas of matrix multiplication become integrals. While for most computer applications, "vectors" refer just to elements of linear spaces, most "vectors" used in models of physical systems are vectors in the original sense of the word, where not only the vector addition and the product with scalars matter, but the products of vectors also have an essential role and their meaning can be best understood in the context of the complete geometric algebra theory.
ReleaseCandidat 5y ago
> A vector is pretty much by definition also a matrix, and there is a standard way to multiply matrices. A standard way to multiply a MxN with a NxK matrix, but none for a 1xN with a 1xN or a Nx1 with a Nx1 matrix - the two possible ways to describe a vector. You have to transpose exactly one of the two vectors. And then you have two possible results, 1xN multiplied with Nx1 yields a scalar (that's actually the 'usual' dot-product/scalar product/whatever you call it) and Nx1 multiplied with 1xN, where the result is a NxN matrix.
- thaumasiotes 5y ago
  Doesn't matter. We already didn't have a unique inverse, but it's perfectly possible to find a left pseudoinverse and a right pseudoinverse, bearing in mind that they're not unique. Though thinking about it more, it seems like the outer-product-inverse of a vector (a) must be unique if it exists; and (b) is highly unlikely to exist. > 1xN multiplied with Nx1 yields a scalar (that's actually the 'usual' dot-product/scalar product/whatever you call it) I'm aware of this, but there are two ways we might conceive of an "inverse": - Since a vector is a matrix, the inverse of a vector might be defined by matrix multiplication, where A is the inverse of B if AB is "the" identity matrix. This is only strictly defined for square matrices, but the pseudoinverse concept extends it to nonsquare matrices. - Or, we could go for a more basic sense of "multiplicative inverse", where the concept is that if AB = C, then B = A⁻¹C. This is what I was thinking of when saying that the concept of an inverse doesn't make sense when multiplication is the inner product - if I give you a vector v, and its inner product with some other vector u, there is no way of recovering what u was.

BlackFly 5y ago

If you are working in ordinary Euclidean space with orthonormal bases, it doesn't do much for you. When you start doing calculus on embedded surfaces (the beginning of differential manifolds) it begins to be more helpful. Since Lie algebras are special types of differential manifolds, you can learn quite a bit by studying the geometry and this quickly leads into gauge theory and modern physics. The Geometry of Physics: An Introduction, by Theodore Frankel does a good job illustrating a lot of aspects of geometry even offering some geometric insight into some classical physics. It is an absolute necessity for general relativity.

contravariant 5y ago
Is there some canonical metric on Lie algebras that makes this possible? Otherwise geometric algebra won't get you far.
- enkimute 5y ago
  Every Lie algebra is a bivector algebra (see 'Lie groups as Spin groups' http://geocalc.clas.asu.edu/pdf/LGasSG.pdf http://geocalc.clas.asu.edu/pdf/LGasSG.pdf). Additionally the GA formalism enables closed form solutions for the exponential map for all bivector algebras. (see 'Graded symmetry Groups' https://www.researchgate.net/publication/353116859_Graded_Symmetry_Groups_Plane_and_Simple/ https://www.researchgate.net/publication/353116859_Graded_Sy...).
Grustaf 5y ago
That books seems very interesting, but is it really about Geometric Algebra? I'm not talking about geometry in general, or differential geometry, which I'm a bit familiar with. And not Algebraic Geometry either, for that matter, which is also a fascinating subject. I mean specifically Geometric Algebra. It sort of seems like a notation, but it has almost a cult like following and perhaps it's more than a notation, is it a theory, a branch of mathematics?
- yuppiemephisto 5y ago
  It really is useful even at the advanced level. Following it far enough leads to the Atiyah-Singer Index Theorem and Hodge Theory. The advantage over the exterior algebra is that you have that and an interior algebra, which leads to many formulas in differential geometry becoming very natural (like Cartan's magic formula).

rsj_hn 5y ago

It's just some of the more advanced theory you'd get from studying modules repackaged a bit. Basically the extra stuff that is usually skipped in first year linear algebra courses are the symmetric and asymmetric (often called exterior) products. These form algebras, of course. The exterior product, or wedge product, has a natural interpretation in terms of signed areas (or volumes) and from this you get the determinant as a volume form. These are the natural generalizations of dot products (inner products) and wedge products (exterior products). You can take a vector and associate it with a 1 form (asymmetric algebra or exterior algebra), and then multiply two vectors to get a 2 form using the standard wedge product, etc. In dimension 3, the space of 2 forms is dual to the space of 1 forms and so you can "multiply" two vectors to get a third vector. That is all that's going on here. Actually a good multi-variable calculus class will cover most of this stuff as you need some motivation for Jacobian volume forms used to calculate areas and volumes under change of basis, and dot/wedge products are useful for generalizations of the Gauss divergence theorem and the generalized fundamental theorem that says the integral over a function, f, on the n-1 dimensional boundary of a shape is the differential of the integral of the shape. Moreover any class on Riemannian geometry will give you all the linear algebra you need as well. One thing I would caution students with is that by using somewhat non-standard jargon they may not understand how to generalize this stuff to n-dimensions, nor will the connections between, say, determinants and wedge-forms be clear, or dot products and angles be fully understood if only the n=3 cases is emphasized. Only in n=3 can you multiply two vectors to get a vector. But fun fact: in dimension 3k you can multiply two k-forms to get a third k-form (as the space of 2k forms is dual to the space of k forms in n=3k). If you think there is this new thing called "geometric algebra" other than usual tensor products, it may not be obvious how things generalize to n != 3.

Grustaf 5y ago
> But no, there is nothing new here beyond marketing. Yes, that's my sense too. Of course cross products, wedge products etc make sense and that's just standard mathematics, but the part that I haven't really seen the point of is to form the algebra where all these forms live side by side. It doesn't seem like a useful "fusing", in the way that say the complex plane is. Of course it's very cool that sub-algebras in 2 or 3-space in GA are isomorphic to the complex plane or even quaternions, but it still feels a bit made up. For a concrete example, one youtuber showed how Maxwells equations simplified to a single equation if you introduce an operator that is a combination of div and curl, and also a new kind of physical entity that combines the electrical and magnetic fields. This is of course cool, but what I want to know is if this new operator makes some physical sense, and if the new multi dimensional field has any physical meaning. If they don't, it just seems like a parlour trick. Not saying they actually don't, but I haven't seen any deeper explanations of it.

nocturnial 5y ago

> Ah, another geometric algebra evangelist? Heh, this made me chuckle and this is reason why I read HN comments. I can't help but to picture someone ringing your doorbell early on an sunday and saying: "Hi! Have you considered inverting vectors?" slams door "Who was it?" "It was just another of those damned geometric algebraists"

Grustaf 5y ago
Haha that reminds me, got to put back that "No beggars or geometric albraists" sign!

LotusFunctor 5y ago

For me, the substantial thing geometric algebra gave me so far was a newfound appreciation of the seemingly disparate systems: tensors, differential forms, matrix algebra, and also a newfound appreciation of stuff like determinants, conjugate elements in group theory, lie groups and lie algebras, etc., because it helps clarify the relationship between them, and as another user here said, you can get propelled up into some pretty advanced stuff later on (said user mentions the Atiyah-Singer Index Theorem and Hodge theory, but caveat: I've only recently started tacking a crack at the latter. I will say that, OTOH, it's pretty nice to be able to see something like the wiki on Clifford Analysis and realize its familiar territory from geometric calculus).