Talk:Geometric algebra

This is maybe too big for a task

I'm pretty sure people will say that, and maybe they're right. But maybe not. I don't think it is much more complicated than say Quaternion type, and in any case it is, from both the programming and mathematical points of view,, quite interesting and worth featuring in Rosetta Code, imho. Please feel free to argue about it.--Grondilu (talk) 22:38, 13 October 2015 (UTC)

It's not clear that we can meaningfully implement anything with infinite dimension - countable or not. At best, we can support a finite subset of such a thing.

More specifically, how would we tell whether an implementation has or has not satisfied that part of the task requirement? --Rdm (talk) 12:36, 14 October 2015 (UTC)

It's infinite in the sense that there is no limit to the number of dimensions. But we only consider vectors that have a finite support. I guess I could mention that, indeed. It's also true that it's not obvious how we can check that an implementation can handle any vector size. I welcome suggestions.--Grondilu (talk) 15:56, 14 October 2015 (UTC)

I would describe that as an arbitrary number of dimensions rather than an infinite number of dimensions (edit: and I see that you have made exactly that change in the task description). But that brings up another issue: ${\begin{array}{c}\forall \mathbf {x} \in {\mathcal {V}},\,\mathbf {x} ^{2}\in \mathbb {R} \end{array}}$ - unless we severely constrain our work, how are we going to verify that an implementation has satisfied this task in that regard? --Rdm (talk) 16:25, 14 October 2015 (UTC)

There is no way to verify this kind of things from just the output. We have to trust the code to do something significant without cheating. What I suggested in the task description was to pick a random vector and check that its square is a Real.--Grondilu (talk) 16:28, 14 October 2015 (UTC)

Ok, so I think what you are saying is roughly that the task should instead be implementing ${\begin{array}{c}\exists \mathbf {x} \in {\mathcal {V}},\,\mathbf {x} ^{2}\in \mathbb {R} \end{array}}$ with the added constraint that ${x}$ must be picked arbitrarily from the space? That's certainly doable... --Rdm (talk) 16:34, 14 October 2015 (UTC)

No, I don't say that. We should keep the axioms as they are and try our best to verify them from a computational point of view.--Grondilu (talk) 16:44, 14 October 2015 (UTC)

After some thought I changed the task requirements and made it demand an implementation of quaternion to demonstrate the solution. Hope it's ok.--Grondilu (talk) 10:37, 17 October 2015 (UTC)

That might be a bit overly constrained? Though, granted, not constrained enough for the "forall" test to be implemented - which, I guess, is why you made the test for correctness be a single case? See also Quaternion type for quaternion implementations... --Rdm (talk) 12:45, 17 October 2015 (UTC)

I see that you posted a solution using the quaternion task. That is obviously not what was intended. You're supposed to implement the geometric algebra and use it to implement quaternions. Implementing quaternions is not the purpose per se.--Grondilu (talk) 16:14, 17 October 2015 (UTC)

I believe I have changed that implementation to satisfy the requirement of the "incorrect" notice. That said, I will also note that I believe this issue could be stated significantly more clearly in the task description with relatively little effort. --Rdm (talk) 17:42, 17 October 2015 (UTC)

I think it is pretty clear already that the quaternion data structure is meant only as a show-case for the use of geometric algebra. Using an implementation of quaternions that does not actually implement geometric algebra can not be satisfactory. If you persist I guess I will have to switch back to verifying axioms.--Grondilu (talk) 17:58, 17 October 2015 (UTC)

Whatever you do, please be sure you understand the requirements stated in the add a task page. You should only expect implementations to satisfy those parts of the task description which are valid for a task description. --Rdm (talk) 18:03, 17 October 2015 (UTC)

You removed the incorrect banner, but didn't you write that the first element of this structure is the real component? If so, e2 defined with 1 0 0 0 is a scalar, not a vector. In any case it is not orthonormal with the e1 and e3. I will add the creation of a function e(n), along with an orthonormality test, as a requirement to clarify that.--Grondilu (talk) 18:12, 17 October 2015 (UTC)

I think you are confused.

So... let's review: a quaternion consists of four orthogonal components. What you term the "scalar part" is a part of an orthonormal basis for quaternions. --Rdm (talk) 18:30, 17 October 2015 (UTC)

A scalar is not orthogonal to a vector.--Grondilu (talk) 18:40, 17 October 2015 (UTC)

Except we are working with quaternions, and the "scalar part of a quaternion" is indeed orthogonal to any "vector part of a quaternion". --Rdm (talk) 18:57, 17 October 2015 (UTC)

I'm not sure what you mean by orthogonal here, because there is no scalar product in the space of quaternions. I did define a scalar product for the purpose of this task, and with this scalar product, a scalar is not orthogonal with a vector, because it commutes with all vectors. For instance 2i + i2 is not 0.--Grondilu (talk) 19:05, 17 October 2015 (UTC)

J's orthonormality expression.

The J expression 0 1 2 3 dot&e"0/0 1 2 3 generates a table of 16 dot products.

Breaking this down:

The left argument is 0 1 2 3, and the right argument is 0 1 2 3. each row corresponds to an item from the left argument (the first row corresponding to the first item from the left argument, the last row corresponding to the last item from the left argument - in other words, rows in the result are in the same order as items in the left argument. Likewise columns in the result correspond to items from the right argument.

In each case, the value in the result is the result of (e left_item) dot (e right_item).

As the result is an identity matrix, we can see that we have an orthonormal basis.

while you're at it, you can clarify the function e and its output for 1, 2, 3.--Grondilu (talk) 19:08, 17 October 2015 (UTC)

A possible source of confusion

The wikipedia page on Clifford algebra states:

The basis elements can be identified with the quaternion basis elements i, j, k as

$i=\mathbf {e} _{2}\mathbf {e} _{3},j=\mathbf {e} _{3}\mathbf {e} _{1},k=\mathbf {e} _{1}\mathbf {e} _{2},$

This has a different parity from the current task requirement for i, j and k. --Rdm (talk) 19:01, 17 October 2015 (UTC)

Well this is not accurate, because with this definition, $ijk=1$, not -1.

$ijk=e_{2}e_{3}e_{3}e_{1}e_{1}e_{2}=e_{2}(e_{3}e_{3})(e_{1}e_{1})e_{2}=e_{2}e_{2}=1$

--Grondilu (talk) 19:12, 17 October 2015 (UTC)