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I ask a previous question which got a really thorough answer.

On the comments it is mentioned that in a geometric algebra you can add blades of different grades.

The goal for me is to understand how to compute these objects, so that is what I am focusing on.

From the prior question I get that a blade of grade $k$ can be represented by an array of length $n \choose k$.

Say we are adding blades of different grades, a scalar and a vector, for example. The scalar is an array of size 1 and the vector an array of size n. What is a potential representation of their sum? is it an array of length n+1?

Makogan
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  • You need a new datatype of size at least $n+1$. A sum of multivectors of different grades is like an unorderded list of grades; you can manipulate each grade separately, and you can extract any grade you wish. – Nicholas Todoroff Oct 31 '22 at 14:04
  • What do you mean by "blade"? They are only mentioned in one comment in the other thread, and in one sentence in the linked wikipedia article. In both cases, as here, the term isn't defined. – Paul Sinclair Oct 31 '22 at 21:18
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    In any case, note in the answer that he says the entire algebra can represented as an array of size $2^n$. In that array, there are $\binom n0 = 1$ indices that correspond to grade $0$ basis multivectors, $\binom n1$ indices that correspond to grade $1$ basis multivectors, ..., $\binom nn$ indices that correspond to grade $n$ basis multivectors. A grade $k$ multivector will only have non-zero entries for the grade $k$ indices. Mixed-grade multivectors will have non-zero entries in more than one grade. – Paul Sinclair Oct 31 '22 at 21:29
  • @PaulSinclair - A blade is usually defined as a multivector that can be represented as (a scalar multiple of) the wedge product of some number of vectors. For example, $e_{1,2}+5e_{1,3}=e_1\wedge(e_2+5e_3)$ is a blade, but $e_{1,2}+e_{3,4}$ is not a blade. (Both of them are grade $2$ multivectors, i.e. bivectors.) Any multivector with grade $0$, $1$, $n-1$, or $n$, is always a blade. And my basis multivectors are also blades; for example $e_{1,2,4}=e_1\wedge e_2\wedge e_4$. – mr_e_man Nov 02 '22 at 17:50
  • @mr_e_man - Thanks. I suspected as much, but wanted confirmation. Then to rephrase my description of your answer, a blade of grade $k$ will have $0$ for all entries in the array except the $\binom nk$ indices of grade $k$. Linear combinations of blades of different grades will have non-zero entries in indices for multiple grades. – Paul Sinclair Nov 02 '22 at 18:58

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