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@ecterrab  Thanks...

snpa 35
August 26 2021
0 0

@ecterrab 

Thanks

@ecterrab  Thans for your advice, a...

snpa 35
August 25 2021
0 0

@ecterrab  Thans for your advice, and here is my question:

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with(Physics)
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Setup(noncommutativeprefix = {x, y})

[noncommutativeprefix = {x, y}]

 (1)
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``
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# If we have known that AntiCommutator(x,x)=0 and AntiCommutator(y,y)=0, and I set algebrarules by
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Setup(algebrarules = {%AntiCommutator(x, x) = 0, %AntiCommutator(y, y) = 0})

[algebrarules = {%AntiCommutator(x, x) = 0, %AntiCommutator(y, y) = 0}]

 (2)
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# From the above algebrarules, MAPLE now treats x and y as "anticommutative" variables.
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``
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GrassmannParity(x), GrassmannParity(y)

1, 1

 (3)
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# Since x and y are GrassmannParity=1 varialbes, we know that x*y=-y*x, thus AntiCommutator(x,y)=x*y+y*x=0.
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``
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# While my question is for two matrices by
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``
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E1 := Matrix([[], [0, 0, 1], []]); E2 := Matrix([[], [], [0, 1, 0]])

Matrix(3, 3, {(1, 1) = 0, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = 0, (2, 3) = 1, (3, 1) = 0, (3, 2) = 0, (3, 3) = 0})

  

Matrix(%id = 18446746504093738878)

 (4)
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# Calculating their anticommutator, we have
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AntiCommutator(E1, E2)

Matrix(%id = 18446746504244204238)

 (5)
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# which is not zero. But we have
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``
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AntiCommutator(E1, E1), AntiCommutator(E2, E2)

Matrix(%id = 18446746504110881838), Matrix(%id = 18446746504110881958)

 (6)
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``
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# As far as I think, the outcome of matrices E1,E2 comparing the outcome of the noncommutative variables x,y, are different.
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# So my problem is: When two variables have AntiCommutator(x,x)=0 and AntiCommutator(y,y)=0, this will not lead to AntiCommutator(x,y)=0.
#

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``
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# If I have made any mistake, please tell me what is going wrong. Thanks a lot !
> 

``


 

Download Problem_on_AntiCommutator.mw

 

 

@Pascal4QM  Thanks for your answer...

snpa 35
August 20 2021
0 0

@Pascal4QM 

Thanks for your answer.

The backgroud goes like:

There is a free Lie algebra  \frak{g} generated by x[1],x[2],...x[n] with some commutation relations, additionally, a Lie superalgebra.

In the case of sl(2|1), e23 and e32 are two generators and their anticommutator is e22+e33 while e23.e23+e23.e23=e32.e32+e32.e32=0.

So take both of above into consideration, if AntiCommutator(x[1],x[1])=0 and AntiCommutator(x[2],x[2])=0, and in some special cases, their realization might be e23 and e32, then there is a certain contradiction in maple.

I thought the reason is that maple naturally take the variable x[1], when AntiCommutator(x[1],x[1])=0, as an odd variable. It may not an odd variable, for example, an generator of a Lie superalgebra.

So maybe I shall not use the "noncommutative variables" to calculate the realization of Lie algebra? Can you offer some good devices that perform well in Lie algebra(Lie bracket) settings?

Thanks a lot!

 

@vv  I finally know some logic of ...

snpa 35
March 17 2019
0 0

@vv 

I finally know some logic of maple's...

is there anyway or any packages to 'define' integral's constant to be zero so that i can do the simplification?

thanks a lot

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