## 35 Reputation

5 years, 197 days

## @ecterrab  Thanks...

Thanks

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 (1)
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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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 (2)
 > # From the above algebrarules, MAPLE now treats x and y as "anticommutative" variables.
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 (3)
 > # 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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 > # While my question is for two matrices by
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 (4)
 > # Calculating their anticommutator, we have
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 (5)
 > # which is not zero. But we have
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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.
 > # 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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 > # If I have made any mistake, please tell me what is going wrong. Thanks a lot !
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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 ...

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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