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The thing is group theory is taught in an incredibly abstract manner, its hard to find any motivating application for it, or any problems it helps us solve. Al

The thing is group theory is taught in an incredibly abstract manner, its hard to find any motivating application for it, or any problems it helps us solve. Also terminology/definitions are vague too, whether a vector has an endpoint or it something unachored in space is itself not clear from many treatments.

gspr 5y ago
> The thing is group theory is taught in an incredibly abstract manner, its hard to find any motivating application for it, or any problems it helps us solve. Mathematics is abstractly defined. But for basic group theory there's a plentitude of very concrete examples to rely on. > Also terminology/definitions are vague too, Absolutely not. There is no vagueness at all! Everything is completely well-defined in most introductory textbooks/courses (or you can even read the precise definitions on Wikipedia, which is often not the case). > whether a vector has an endpoint or it something unachored in space is itself not clear from many treatments. Vectors do not have endpoints. Vectors are not anchored. Vectors are elements of vector spaces. Vector spaces are completely clearly defined.
- jcora 5y ago
  Sounds snarky but completely correct. Parent should look for a more diverse set of examples for vector spaces. In fact sounds like a good linear algebra course would be a priority over group theory
- thaumasiotes 5y ago
  > Vectors do not have endpoints. Vectors are not anchored. Vectors are elements of vector spaces. Vector spaces are completely clearly defined. Ehh.... compare this text from the wikipedia article on affine spaces: > In an affine space, there is no distinguished point that serves as an origin. Hence, no vector has a fixed origin and no vector can be uniquely associated to a point. In an affine space, there are instead displacement vectors, also called translation vectors or simply translations, between two points of the space. If you're working with vectors, you're generally working with them as points, not as values that happen to obey the axioms defining a vector space. Those vectors are anchored, and the anchoring is so deeply embedded in the concept that there's a separate concept, affine spaces, specifically devoted to the question of "what if we had things that were like vectors, except without being anchored to a particular point in space?"
  - kmill 5y ago
    Generalizing this, in the theory of differentiable manifolds, a vector (well, a "tangent vector") can be thought of as being an arrow rooted at a specific point. In Euclidean space (affine spaces), you can always translate all vectors to an origin, so it's safe to ignore a vector's root, but in general there's not a canonical way to translate vectors. The issue is essentially curvature -- for example, if you took a tangent vector at the north pole of a sphere, dragged it down to the equator, dragged it along the equator for a bit, then dragged it back up to the north pole, the vector would have rotated. What this shows is that vectors at different points are not mutually comparable without more structure. I swept a detail under the rug, which is that the vectors are infinitesimal arrows, so their tip is not actually another point of the manifold. In an affine space, vectors can be regarded as non-infinitesimal arrows -- the arrows that you always find textbook illustrations on vector arithmetic. The arrows with a common root form a vector space.
    - gspr 5y ago
      Right, but the individual vector spaces you speak of there are the tangent spaces of a manifold M. There's a vector space TpM attached to M at a point p. And the point you make is that there isn't necessarily a canonical way to relate vectors in TpM to vectors in TqM for p≠q. In this sense, we may well say that "a vector in TpM is anchored at p, and a vector in TqM is anchored at Q". I agree with that, obviously. However, within a fixed vector space, of which such a TpM is merely one example, there is no notion of vectors being "anchored".
      - kmill 5y ago
        Given the definitions in a textbook on smooth manifolds, I agree with you -- the modern notion of a vector space has no notion of root or anchor since these words have no associated definitions. It's interesting to think about how we came to the modern viewpoint. Some of the underlying context behind my comment is that long ago manifolds were concretely subsets of R^n that could be locally parameterized. The tangent space at a point was the affine subspace of R^n that was tangent to the manifold there, and you could regard a tangent vector as an arrow rooted at that point inside that subspace. Even though we have our clean modern notion of an abstract vector space, somehow these old intuitions linger on in other disciplines In math we have a precise definition for a vector (an element of a vector space, full stop), but I don't think experts have the authority to be prescriptive about jargon outside their discipline. Of course, a student should be willing to drop preconceptions and to absorb correct usage.
- billfruit 5y ago
  What I was trying to tell was that even a fundamental concept such as vector isn't clearly defined in undergraduate level treatments. And consider that fields like Physics, EE, Aeronautics and Computer Graphics use such concepts, often use different definitions of the same thing even.
  - gspr 5y ago
    > What I was trying to tell was that even a fundamental concept such as vector isn't clearly defined in undergraduate level treatments. Of course it is. I'm sure there are instances where it isn't, but it also mostly is.
immmmmm 5y ago
it's actually more group representation theory you'll need: the rotation group has an infinite number of representations acting on different vectors spaces, rotation of an electron in 3d is SU(2) which maps to SO(3), the rotation of vectors.
swiley 5y ago
That's an odd complaint to hear on a technology centered forum. The most obvious application for algebra should be thinking about data structures with their operations. If you can't be sure about a method on a class being valid in terms of the contracts for that class (IE being a closed operation) then the method can't be public (usually an idea taught in "intro to OOP" style classes although with other words unless you're reading something like SICP.)
klodolph 5y ago
Some of the motivating examples are hard to understand, unfortunately. You can't do quantum mechanics without group theory, and if you can't do quantum mechanics, it will be much harder to understand how half the instrumentation in your chemistry lab works.
FabHK 5y ago
> taught in an incredibly abstract manner And that’s a shame, because a lot of group theory originated in concrete, tangible problems. Check out the book Visual Group Theory by Nathan Carter, discussed (a bit) here: https://news.ycombinator.com/item?id=11745486 https://news.ycombinator.com/item?id=11745486 https://news.ycombinator.com/item?id=16345844 https://news.ycombinator.com/item?id=16345844

gspr 5y ago

> The thing is group theory is taught in an incredibly abstract manner, its hard to find any motivating application for it, or any problems it helps us solve. Mathematics is abstractly defined. But for basic group theory there's a plentitude of very concrete examples to rely on. > Also terminology/definitions are vague too, Absolutely not. There is no vagueness at all! Everything is completely well-defined in most introductory textbooks/courses (or you can even read the precise definitions on Wikipedia, which is often not the case). > whether a vector has an endpoint or it something unachored in space is itself not clear from many treatments. Vectors do not have endpoints. Vectors are not anchored. Vectors are elements of vector spaces. Vector spaces are completely clearly defined.

jcora 5y ago
Sounds snarky but completely correct. Parent should look for a more diverse set of examples for vector spaces. In fact sounds like a good linear algebra course would be a priority over group theory
thaumasiotes 5y ago
> Vectors do not have endpoints. Vectors are not anchored. Vectors are elements of vector spaces. Vector spaces are completely clearly defined. Ehh.... compare this text from the wikipedia article on affine spaces: > In an affine space, there is no distinguished point that serves as an origin. Hence, no vector has a fixed origin and no vector can be uniquely associated to a point. In an affine space, there are instead displacement vectors, also called translation vectors or simply translations, between two points of the space. If you're working with vectors, you're generally working with them as points, not as values that happen to obey the axioms defining a vector space. Those vectors are anchored, and the anchoring is so deeply embedded in the concept that there's a separate concept, affine spaces, specifically devoted to the question of "what if we had things that were like vectors, except without being anchored to a particular point in space?"
- kmill 5y ago
  Generalizing this, in the theory of differentiable manifolds, a vector (well, a "tangent vector") can be thought of as being an arrow rooted at a specific point. In Euclidean space (affine spaces), you can always translate all vectors to an origin, so it's safe to ignore a vector's root, but in general there's not a canonical way to translate vectors. The issue is essentially curvature -- for example, if you took a tangent vector at the north pole of a sphere, dragged it down to the equator, dragged it along the equator for a bit, then dragged it back up to the north pole, the vector would have rotated. What this shows is that vectors at different points are not mutually comparable without more structure. I swept a detail under the rug, which is that the vectors are infinitesimal arrows, so their tip is not actually another point of the manifold. In an affine space, vectors can be regarded as non-infinitesimal arrows -- the arrows that you always find textbook illustrations on vector arithmetic. The arrows with a common root form a vector space.
  - gspr 5y ago
    Right, but the individual vector spaces you speak of there are the tangent spaces of a manifold M. There's a vector space TpM attached to M at a point p. And the point you make is that there isn't necessarily a canonical way to relate vectors in TpM to vectors in TqM for p≠q. In this sense, we may well say that "a vector in TpM is anchored at p, and a vector in TqM is anchored at Q". I agree with that, obviously. However, within a fixed vector space, of which such a TpM is merely one example, there is no notion of vectors being "anchored".
    - kmill 5y ago
      Given the definitions in a textbook on smooth manifolds, I agree with you -- the modern notion of a vector space has no notion of root or anchor since these words have no associated definitions. It's interesting to think about how we came to the modern viewpoint. Some of the underlying context behind my comment is that long ago manifolds were concretely subsets of R^n that could be locally parameterized. The tangent space at a point was the affine subspace of R^n that was tangent to the manifold there, and you could regard a tangent vector as an arrow rooted at that point inside that subspace. Even though we have our clean modern notion of an abstract vector space, somehow these old intuitions linger on in other disciplines In math we have a precise definition for a vector (an element of a vector space, full stop), but I don't think experts have the authority to be prescriptive about jargon outside their discipline. Of course, a student should be willing to drop preconceptions and to absorb correct usage.
billfruit 5y ago
What I was trying to tell was that even a fundamental concept such as vector isn't clearly defined in undergraduate level treatments. And consider that fields like Physics, EE, Aeronautics and Computer Graphics use such concepts, often use different definitions of the same thing even.
- gspr 5y ago
  > What I was trying to tell was that even a fundamental concept such as vector isn't clearly defined in undergraduate level treatments. Of course it is. I'm sure there are instances where it isn't, but it also mostly is.

immmmmm 5y ago

it's actually more group representation theory you'll need: the rotation group has an infinite number of representations acting on different vectors spaces, rotation of an electron in 3d is SU(2) which maps to SO(3), the rotation of vectors.

swiley 5y ago

That's an odd complaint to hear on a technology centered forum. The most obvious application for algebra should be thinking about data structures with their operations. If you can't be sure about a method on a class being valid in terms of the contracts for that class (IE being a closed operation) then the method can't be public (usually an idea taught in "intro to OOP" style classes although with other words unless you're reading something like SICP.)

klodolph 5y ago

Some of the motivating examples are hard to understand, unfortunately. You can't do quantum mechanics without group theory, and if you can't do quantum mechanics, it will be much harder to understand how half the instrumentation in your chemistry lab works.

FabHK 5y ago

> taught in an incredibly abstract manner And that’s a shame, because a lot of group theory originated in concrete, tangible problems. Check out the book Visual Group Theory by Nathan Carter, discussed (a bit) here: https://news.ycombinator.com/item?id=11745486 https://news.ycombinator.com/item?id=11745486 https://news.ycombinator.com/item?id=16345844 https://news.ycombinator.com/item?id=16345844