Anyone familiar with Peano' (or Grassman) algebra?

[enigma]

I'm an engineer, not a mathematician...

the professor has assigned several proof questions, and I'm having difficulty answering them...

(This may need to be moved to homework help, but the topic is unusual, so I thought I'd get better response here)

Terminology: v is a join operation, ^ is a meet operation D[] is a bracket operation

Example:

Show that when:

[tex] A = a_1 \vee a_2, B = b_1 \vee b_2, \ and\ n = 2 [/tex]

then

[tex] A \wedge B = -D[a_2 , b_1 , b_2] a_1 + D[a_1 , b_1 , b_2] a_2 [/tex]
[tex]= D[a_1 , a_2 , b_2] b_1 - D[a_1 , a_2 , b_1] b_2 [/tex]

If the meet is zero, then assuming that the sets [itex] (a_1, a_2) [/itex] and [itex] (b_1, b_2) [/itex] are both independent, the four brackets must be zero. Show that, in this case, the two subspaces [itex] A_s[/itex] and [itex] B_s [/itex] are the same.

Now, proving the first part is fairly simple: just run though the definition of the meet.

I'm having difficulty with the second part. I thought that the definition of the meet is the intersection of the two subspaces. If the subspaces are the equal, then wouldn't the meet be either A_s or B_s?

 

[enigma]

Answering my own question somewhat:

OK. If the sets are independant, and [itex] \alpha a_1 + \beta a_2 = 0[/itex] , then [itex] \alpha \ and \ \beta[/itex] need to be zero because independancy implies [itex]a_1 \ and \ a_2[/itex] are not equal.

If [itex] D[a_i, b_i, b_j] \ and \ D[a_i, a_j, b_i] [/itex] are zero for any combination of i and j, that means that the two subspaces A and B are linearly dependant for any combination of bases. This must mean that they are the same subspace.

Is that correct?

 

[Janitor]

Pardon my ignorance; I have barely ever done any reading on Grassman algebra. I probably won't be able to help you. But what is the definition of n?

 

[enigma]

n is the number of dimensions in addition to the projective plane: PA[Rⁿ⁺¹], where PA represents peano algebra.

so, the dimensions of a two dimensional projective space is PA[R²⁺¹] corresponding to e₀ e₁ and e₂ as the three coordinates, with e₁ being the x direction, e₂ being the y direction and e₀ being the perspective direction.

*I hope I'm not botching up my terminology here...*

 

[Janitor]

Ummm

I'll get back to you in a few weeks on this.

 

[Janitor]

You must know the answer by now! If you have time, could you sketch it out for us? If you're too busy, that's understandable.