Orientation in five dimensions

by Thomas   Last Updated April 25, 2018 17:20 PM

I was wondering what the orientation of axes would be in five dimensions.

In 3D, a right-handed orientation means $\vec{x}\times \vec{y}=\vec{z},\space\vec{y}\times \vec{z}=\vec{x},\space\vec{z}\times \vec{x}=\vec{y}$.

Now, I'm trying to understand how this would work in 5D space, or 4D for that matter.

I think it cannot be said that for the vectors $\vec{x},\vec{y},\vec{z},\vec{w},\vec{v}$ in $\mathbf{R}^5$ that $\vec{y}\times \vec{z}=\vec{w}$ instead? This because in $\mathbf{R}^n$, $n$ needs to be $2^k$ for it to work, hence there is a seven-dimensional cross product.

My question would be, that for a five dimensional space, how would the orientation of the axes be? Or would it simply be impossible or incomprehensible?

Answers 2

You no longer have the cross product, as that is a construction that exists only in 3 or 7 dimensions. You can derive the axes in many ways, though. Take a vector $\vec{v}\in\mathbb{R}^n$. Let us pretend that $\vec{v}$ is in fact a row of a matrix, $A$, and let's further pretend that $A$ has no other rows (that is, $A$ is just $\vec{v}$). Then the null space of this matrix is exactly those vectors that are perpendicular to its only row. Finding the null space isn't terribly hard in principle, and it would yield $n-1$ orthogonal vectors that might form your axes. This isn't just one such way to form these axes, though, as any $n$ independent vectors would do it.

We might call them "perpendicular", but be careful about how you visualize that. When we say the word "perpendicular" we tend to mean "we can draw them at right angles". However, we can't really conceive of higher than 3 dimensional spaces, so we need to abstract the idea of "perpendicular" a bit. I'm assuming you are familiar with the dot-product, if not just ask.

In this sense, two vectors are perpendicular if their dot product is zero. We can't draw it, but we can verify that the vectors are perpendicular, if you required your space to have strictly perpendicular axes.

Michael Stachowsky
Michael Stachowsky
April 17, 2018 13:38 PM

You can calculate perpendicular vector $v$ to the 4D hyperplane spanned by four orthogonal vectors $x,y,z,w$ in similar way like for 3D with use of pseudo-determinant formula:

$v=\det \begin {bmatrix} \mathbf i & \mathbf j & \mathbf k & \mathbf l & \mathbf m \\ x_1 & x_2 & x_3 & x_4 & x_5 \\ y_1 & y_2 & y_3 & y_4 & y_5 \\ z_1 & z_2 & z_3 & z_4 & z_5 \\ w_1 & w_2 & w_3 & w_4 & w_5 \\ \end{bmatrix} $, where $\mathbf i , \mathbf j , \mathbf k , \mathbf l , \mathbf m $ are versors of standard basis for 5D.

I would denote such operation as $v=X(x,y,z,w)$.
In 3D adjusted formula would give cross product $z=X(x,y)= x \times y$.
Notice that like in 3D interchanging two rows gives opposite direction of the vector $v$.

April 17, 2018 13:54 PM

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