# Why do odd dimensions and even dimensions behave differently?

It is well known that odd and even dimensions work differently.

Is there some unifying reason for these differences? Why is dimensional parity so important?

• Some of these items (e.g., non-vanishing fields on the sphere) can be explained by noting that an even-dimensional real vector space may be viewed as a complex vector space. – Andrew D. Hwang Dec 5 '13 at 11:43
• You may enjoy some of the answers here: mathoverflow.net/questions/5372/dimension-leaps – Sammy Black Dec 14 '13 at 4:52

For real vector spaces, i.e. $\mathbb{R}^N$, one important difference between odd and even $N$ is that every real polynomial with odd order has a real zero, while there are real polynomials with even order that only have complex zeros.

Thus, a real invertible $N\times N$ matrix always has a non-zero eigenvector if $N$ is odd (because the characteristic polynomial has order $N$, and by the above has a zero). For even $N$, however, there are invertible real matrices with no non-zero eigenvector.

This means that in even-dimensional spaces, there are invertible linear mappings without an invariant one-dimensional subspace, while in odd-dimensional spaces, invertible linear mappings always have a one-dimensional invariant subspace. If you look only at rotations, you get that rotations in an odd-dimensional space always keep at least a line fixed, while in an even-dimensional space they do not.

That's a pretty huge geometric difference between odd- and even-dimensional spaces, that should explain a least some of the items in the question.

This feels more like a big-list/community-wiki question rather than a bounty problem... Here are two examples:

i) An even-dimensional space supports symplectic forms; an odd-dimensional one does not. [Symplectic = bilinear, alternating, and nondegenerate; bilinear means it corresponds to some $n\times n$ matrix $A$, and then alternating means $A$ is skew-symmetric and nondegenerate means $\det A \neq 0$. For skew-symmetric $A$ we have $\det A = \det (-A^*) = (-1)^n \det(A^*) = (-1)^n \det A$, so for $n$ odd the determinant must vanish; while for even $n$ it's easy to construct nondegenerate examples, such as the block matrix $({\phantom-0 \; I \atop -I \; 0})$ where $I$ is the identity matrix of order $n/2$.]

ii) The Euler characteristic of a sphere of dimension $n$ is $1 + (-1)^n$ which is zero iff $n$ is odd. So you can comb an even-dimensional coconut (whose surface is an odd-dimensional sphere) but not an odd-dimensional one. [To comb the unit ball in ${\bf R}^{2m}$, identify ${\bf R}^{2m}$ with ${\bf C}^m$, and associate to each unit vector $v$ the tangent vector in the direction $iv$, which is orthogonal to $v$.]

[Added later: (i) and (ii) are related because the block matrix $({\phantom-0 \; I \atop -I \; 0})$ is rotation by $i$. But I see that the OP already mentioned the "hairy ball theorem" and linked to the same Wikipedia page. So I offer a third example.]

iii) The real orthogonal groups of even and odd order behave differently as Lie groups: $SO_{2m+1}$ (for $m \geq 1$) is simple of type $B_m$, while $SO_{2m}$ ($m \geq 2$) has nontrivial center $\{\pm 1\}$ and its quotient by the center is simple once $m > 2$ and of type $D_m$.