- #1

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## Main Question or Discussion Point

At high school age I had trouble with complex numbers, because there was no rigor definition given to them, but instead only the property [itex]i^2=-1[/itex], and then we were supposed to calculate with it. This lead to somewhat mystical interpretations of imaginary unit sometimes, until I figured out the definition of complex numbers as [itex]\mathbb{R}^2[/itex] with given multiplication rule.

Now I'm having precisely the same problem with Grassmann numbers. I have often encountered "definitions" where the property [itex]xy=-yx[/itex] is given, but nothing more precise about what the numbers actually are. I see it is easy to define an algebra where there is a finite amount of Grassmann variables, but I'm not sure this is satisfactory always. In physics it seems to be, that for example entire complex field [itex]\mathbb{C}[/itex] can be merely promoted to become Grassmann algebra.

Here's my attempt to make Grassmann multiplication onto [itex]\mathbb{R}[/itex]:

We first identify [itex]\mathbb{R}[/itex] with [itex]\mathbb{R}\times\{0\}\subset\mathbb{R}^2[/itex], and then define a multiplication [itex]*:\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}^2[/itex] as follows.

For all [itex]x\in\mathbb{R}[/itex], [itex](x,0)*(x,0)=(0,0)[/itex].

If [itex]0<x<x'[/itex], then [itex](x,0)*(x',0)=(0,xx')[/itex] and [itex](x',0)*(x,0)=(0,-xx')[/itex].

If [itex]x<0[/itex] and [itex]0<x'[/itex], then [itex](x,0)*(x',0)=-(|x|,0)*(x',0)[/itex].

If [itex]x,x'<0[/itex], then [itex](x,0)*(x',0)=(|x|,0)*(|x'|,0)[/itex].

For all [itex](x,y),(x',y')\in\mathbb{R}^2[/itex], [itex](x,y)*(x',y')=(x,0)*(x',0)[/itex].

I think if one wants an anti-commuting multiplication on [itex]\mathbb{R}[/itex], it is necessary to add one dimension like this. You cannot have all products inside the original space. Now this [itex]*[/itex] should be a proper multiplication, with the desired anti-commuting property. Or does there seem to be problems with this definition?

Is there other definitions that would be equivalent with this? Or is there other definitions, which are not equivalent with this?

One thing that disturbs me is that I'm not sure how to do the same thing with [itex]\mathbb{C}[/itex], because there is not natural order relation [itex]<[/itex], so precisely the same definition wouldn't work.

Now I'm having precisely the same problem with Grassmann numbers. I have often encountered "definitions" where the property [itex]xy=-yx[/itex] is given, but nothing more precise about what the numbers actually are. I see it is easy to define an algebra where there is a finite amount of Grassmann variables, but I'm not sure this is satisfactory always. In physics it seems to be, that for example entire complex field [itex]\mathbb{C}[/itex] can be merely promoted to become Grassmann algebra.

Here's my attempt to make Grassmann multiplication onto [itex]\mathbb{R}[/itex]:

We first identify [itex]\mathbb{R}[/itex] with [itex]\mathbb{R}\times\{0\}\subset\mathbb{R}^2[/itex], and then define a multiplication [itex]*:\mathbb{R}^2\times\mathbb{R}^2\to\mathbb{R}^2[/itex] as follows.

For all [itex]x\in\mathbb{R}[/itex], [itex](x,0)*(x,0)=(0,0)[/itex].

If [itex]0<x<x'[/itex], then [itex](x,0)*(x',0)=(0,xx')[/itex] and [itex](x',0)*(x,0)=(0,-xx')[/itex].

If [itex]x<0[/itex] and [itex]0<x'[/itex], then [itex](x,0)*(x',0)=-(|x|,0)*(x',0)[/itex].

If [itex]x,x'<0[/itex], then [itex](x,0)*(x',0)=(|x|,0)*(|x'|,0)[/itex].

For all [itex](x,y),(x',y')\in\mathbb{R}^2[/itex], [itex](x,y)*(x',y')=(x,0)*(x',0)[/itex].

I think if one wants an anti-commuting multiplication on [itex]\mathbb{R}[/itex], it is necessary to add one dimension like this. You cannot have all products inside the original space. Now this [itex]*[/itex] should be a proper multiplication, with the desired anti-commuting property. Or does there seem to be problems with this definition?

Is there other definitions that would be equivalent with this? Or is there other definitions, which are not equivalent with this?

One thing that disturbs me is that I'm not sure how to do the same thing with [itex]\mathbb{C}[/itex], because there is not natural order relation [itex]<[/itex], so precisely the same definition wouldn't work.