Talk:Geometric algebra: Difference between revisions
→A possible source of confusion
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::::::It's not consistent if i, j and k are not orthogonal to each other, which is the case for the javascript implementation (and I do not have a working copy of perl6, so I can't test whether that implementation is invalid or not). It might be that i, j and k are not orthogonal for any implementation which satisfies the constraints in the task description - I suspect that that is the case, but I need to think a bit more to see if I can either (a) find an interpretation of the task description which is internally consistent, or (b) find a way of showing a contradiction. --[[User:Rdm|Rdm]] ([[User talk:Rdm|talk]]) 22:01, 17 October 2015 (UTC)
:::::::i, j and k are not vectors in geometric algebra. They are bivectors. Orthogonality does not apply to them. The formula I gave for the scalar product only stands for vectors (thus the boldface notation which often means "vectors"). Also I've never written they should be orthogonal. It's the e basis that is, not the i, j, k.--[[User:Grondilu|Grondilu]] ([[User talk:Grondilu|talk]]) 22:24, 17 October 2015 (UTC)
::::::::We might still have an issue here. I think I can agree that I, j, and k are not vectors from the orthogonal basis e. But claiming that they are "not vectors" without qualification seems both sloppy and misleading. Needs more exposition, I guess... --[[User:Rdm|Rdm]] ([[User talk:Rdm|talk]]) 18:02, 19 October 2015 (UTC)
== "Orthonormal basis" ==
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