- What is the degree of the antipodal map on the \(n\)-sphere?
(No justification required)
Define a CW complex homeomorphic to the real projective \(n\dash\)space \(\RP^n\).
Let \(\pi : \RP^n \to X\) be a covering map. Show that if \(n\) is even, \(\pi\) is a homeomorphism.
Solution
1. The degree of the antipodal map \(a:S^n\to S^n\), \(a(x)=-x\), is\[\boxed{\deg a=(-1)^{n+1}.}\]
The antipodal map is the restriction of the linear map \(-I\) on \(\mathbb R^{n+1}\), whose determinant is \((-1)^{n+1}\).
2. A CW structure on \(\mathbb{RP}^n\) has one cell \(e^k\) in each dimension \(0\le k\le n\).
Use the filtration \[\mathbb{RP}^0\subset\mathbb{RP}^1\subset\cdots\subset\mathbb{RP}^n,\] where \(\mathbb{RP}^k\) is obtained from \(\mathbb{RP}^{k-1}\) by attaching a \(k\)-cell; the attaching map is the antipodal quotient \(S^{k-1}\to\mathbb{RP}^{k-1}\) on the boundary of the upper hemisphere.
3. Suppose now that \(n\) is even and \(\pi:\mathbb{RP}^n\to X\) is a covering map. Its number of sheets \(d\) is finite.
The total space is compact. A fiber is a closed discrete subspace of the compact space \(\mathbb{RP}^n\), hence finite. Since the base is connected, all fibers have the same finite cardinality \(d\).
4. Euler characteristic multiplicativity gives\[1=\chi(\mathbb{RP}^n)=d\,\chi(X).\]
For even \(n\), the one-cell-per-dimension CW structure gives \(\chi(\mathbb{RP}^n)=1\). Euler characteristic multiplies under finite coverings.
5. Hence \(d=1\), so \(\pi\) is a homeomorphism.
Both \(d\) and \(\chi(X)\) are integers and \(d>0\), so \(1=d\chi(X)\) forces \(d=1\). A one-sheeted covering is a homeomorphism.