Intuition for orientation of tri-vectors in geometric algebra

Question

I am learning geometric algebra from the MacDonald textbook and it states that the outer product is associative. Letting $\bf{u}$, $\bf{v}$, and $\bf{w}$ be vectors

$$\bf{u} \wedge \bf{v} \wedge \bf{w}= (\bf{u} \wedge \bf{v}) \wedge \bf{w} =\bf{u} \wedge (\bf{v} \wedge \bf{w}).$$

But the book describes the volume created by $(\bf{u} \wedge \bf{v}) \wedge \bf{w}$ has orientation given by $\bf{u} \wedge \bf{v}$, while $\bf{u} \wedge (\bf{v} \wedge \bf{w})$ has orientation given by $\bf{v} \wedge \bf{w}$.

I understand that they are the same volume, but what I'm not getting is the understanding of what the orientation of an oriented volume actually is. Is it saying that $\bf{u} \wedge \bf{v}$ and $\bf{v} \wedge \bf{w}$ have the same orientation even though they are not in parallel planes?

Also, any tips on gaining an overall intuition of orientation of higher dimensional objects would be greatly appreciated.

UPDATE

An updated version of the section in question has been posted here:

http://www.faculty.luther.edu/~macdonal/laga/LAGA%20Section%205.2.pdf

This rewritten section provides a different explanation than above.

Answer 1

I'm not familiar with MacDonalds book, but this does not sound right to me; the orientation of a volume element $u\wedge v \wedge w$ should depend upon all three factors and not just two of them.

The outer product is otherwise known as the wedge or exterior product. It is a signed version of the tensor product, hence used when orientation matters.

They generalise vectors; recall that vectors have direction and magnitude. The outer product of two vectors give a '2d vector', an orientated area element, this ought to be thought of as the generalisation of direction to higher dimensions, where instead of working in higher dimensions per se - for example in a 10d or 20d space - we are working with higher dimensional vectors; the generalisation of magnitude then is the area.

The generalisation to 3d vectors is then obvious: the outer product of three vectors gives an orientated volume element and the magnitude is now the volume.

This explains why the outer product of a vector by itself always vanishes; this is because the area element degenerates into a line element, and hence it's area vanishes; likewise its volume.

The arithmetic of the outer product shows that a generalisation of the triangle law still holds, so long as the two area elements in question have an edge of common magnitude; it's also important to note that the multi-linearity of the outer product means that we can rescale the area element in such a way that it's total area does not change; for example we can magnify one edge by a scale factor $k$ so long as we reduce the other side by $k$ - this has no parallel in the arithmetic of vectors.

Answer 2

The orientation of $\bf{u} \wedge \bf{v}$ is given by the orientation of $\bf{u}$ on the boundary of $\bf{u} \wedge \bf{v}$. See Fig. 5.5 in my book and the accompanying text.

Similarly, the orientation of $\bf{u} \wedge \bf{v} \wedge \bf{w}$ is given by the orientation of $\bf{u} \wedge \bf{v}$ (or $\bf{v} \wedge \bf{w}$ or $\bf{w} \wedge \bf{u}$) on the boundary of $\bf{u} \wedge \bf{v} \wedge \bf{w}$. All are clockwise.

Perhaps this is a better explanation than in my book.