## Transcript of 005.AVI (Lecture 5)

By Eckhard M.S. Hitzer, University of Fukui, Prof. Ryosuke Nagaoka (University of the Air, Japan)

So now please have a look at the computer screen. And here you see a map f of the space R n (upper index n!), which is a linear space over the real numbers, and it is mapped into itself, therefore here again into R n (upper index n!).

Here you see R n for the original and R n for the image (space). So first the vector a one (a1) is mapped into f of a one [f(a1)]. Than another vector a two (a2) is mapped into f of a two [f(a2)] and a third vector a three (a3) is mapped into f of a three [f(a3)]. So in total this cube is mapped into this parallelopiped.

... Prof. Nagaoka ...

Yes. With the help of the linear map of a vector space into itself, we can now define also a linear map of the geometric algebra of this vector space into itself. So let us look at the next computer graphics.

Here we have the geometric algebra of the space R n (upper index n!) and the geometric algebra has the lower index n, to distinguish it from the vector space it belongs to. And here you see a vector and the map which we had before mapping it into f of a one [f(a1)], and as you know a two (a2) is mapped into f of a two [f(a2)]. And now in the geometric algebra R lower index n we have not only vectors but we have bivectors, and so this bivector is a one hat a two (a1 ^ a2). It is an outer product but it is not the same outer product, which one is used to only in three dimensions. This outer product really produces this oriented area in space itself. And the outermorphism is mapping this oriented area into the outer product of the two images of the side vectors here. And we call this f underbar, so it is the outermorphism, which maps a one hat a two (a1 ^ a2) into f of a one hat f of a two [f(a1) ^f(a2)] here. So this is precisely the definition of f underbar of a one hat a two (a1 ^ a2), the image of the bivector under the outermorphism.

Then we have a third vector a three (a3), it is mapped into f of a three [f(a3)]. And here we have a volume element a one hat a two hat a three (a1 ^ a2 ^ a3) and we have a corresponding volume element in the image space f of a1 hat f of a2 hat f of a3 [f(a1) ^f(a2) ^ f(a3)]. And the outermorphism is mapping the original volume element into the corresponding image volume element. And again the symbol is f underbar for outermorphism. So f underbar of the original volume element a one hat a two hat a three (a1 ^ a2 ^ a3) equals this corresponding volume element in the image space.

... Prof. Nagaoka ...

Yes. Next I will explain how with the help of the outermorphism we can make a very nice and very straight-forward definition of the determinant. Please look at the computer screen.

Now we just restrict ourselves to three dimensions, but you can easily generalize it to n dimensions. So here we have again the space for the original and for the image. And so R lower index three means the geometric algebra of the three dimensional vector space. We have a vector e one (e1), which is mapped into f of e one [f(e1)], another unit vector e two (e2) mapped into f of e2 [f(e2)] and a third unit vector e three (e3) mapped into f of e three [f(e3)]. And they are all perpendicular, so they form an orthonormal coordinate system, so to say.

Now this makes just a unit volume element cube. And it is mapped by the outermorphism into this parallelopiped. Here we use the symbol I for the unit cube. And here this is the outermorphism mapping into the parallelopiped. And this is f underbar of I. And we can now divide the image volume f underbar of I into the same units, which we have on the left side. (And) So the result is a scalar, because volume divided by volume is just a scalar. And this is what is called the determinant of f.

... Prof. Nagaoka ...

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Soli Deo Gloria. Created by Eckhard Hitzer (Fukui).