Problem: Define the signature of a matrix. Hence, state and prove Sylvester’s law of inertia.
Solution: The signature of an \(N\times N\) matrix \(A\) is a \(3\)-tuple \((N_+,N_-,N_0)\) where \(N_+\) is the number of positive eigenvalues of \(A\) (including multiplicity), \(N_-\) is the number of negative eigenvalue of \(A\) (including multiplicity), and \(N_0=\text{dim}\ker A\) is the multiplicity of the zero eigenvalue; thus, \(N_++N_-+N_0=N\).
Let \(A,B\) be real symmetric matrices. Then Sylvester’s law of inertia asserts that \(A\) and \(B\) are congruent matrices iff they have the same signature (which sometimes is called “inertia” because of this invariance under congruence, hence the name).
Proof: It’s easy to see that the nullity \(N_0\) is preserved by congruence transformations. If one can can show that \(N_+\) is also preserved, then it implies \(N_-\) is conserved by virtue of \(N_++N_-+N_0=N\). To show this, the idea is to prove by contradiction, assuming \(N_+\) is not preserved which by dimension counting would imply a non-zero vector living in the intersection of the subspace spanned by the positive-eigenvalue eigenvectors of \(A\) and the congruence-transformed subspace spanned by the non-positive-eigenvalue eigenvectors of \(B\).
Since any real, symmetric matrix \(A\) is isomorphic to a real quadratic form \(Q(\mathbf x):=\mathbf x^TA\mathbf x\), the concepts of signature and Sylvester’s law of inertia can also be reformulated in the language of quadratic forms rather than real symmetric matrices.
Problem: Let \(X\) be a smooth manifold. Explain what it means to place the additional structure of a pseudo-Riemannian metric \(g\) on \(X\). Then explain how Riemannian and Lorentzian geometry are special cases of pseudo-Riemannian geometry.
Solution: A Riemannian metric \(g\) on \(X\) is a type \((0,2)\) tensor field that defines an inner product \(g_x:T_x(X)^2\to\mathbf R\) at the tangent space \(T_x(X)\) of each point \(x\in X\). The “tensor field” part is sometimes rephrased as saying that \(g_x\) is a bilinear form. Moreover, to flesh out the usual axioms of a real inner product space, the Riemannian metric tensor \(g_x\) at each \(x\in X\) must be symmetric \(g_x(v_x,v’_x)=g_x(v’_x,v_x)\) and positive-definite \(v_x\neq 0\Rightarrow g_x(v_x,v_x)>0\).
A pseudo-Riemannian metric relaxes the positive-definite requirement of a Riemannian metric, instead merely requiring non-degeneracy (i.e. at each point \(x\in X\), the zero vector \(0\) is the only vector orthogonal \(g_x(0,v_x)=0\) to all \(v_x\in T_x(X)\)). More concisely, what this is saying is that \(N_0=0\), so the signature of a pseudo-Riemannian metric may be thought of as a pair \((N_+,N_-)\).
Thus, Riemannian metrics are the special subset of pseudo-Riemannian metrics for which \((N_+,N_-)=(N,0)\). Meanwhile, Lorentzian metrics are another special subset of pseudo-Riemannian metrics for which \((N_+,N_-)=(1,N-1)\) (assuming the particle physics/QFT convention). Typically, spacetime \(X\) has dimension \(N=4\), so e.g. the Minkowski metric is said to have signature \((1,3):=(+,-,-,-)\).
Problem: Let \((X,g)\) be a pseudo-Riemannian manifold. Often, one writes the general expression for an infinitesimal line element on \(X\) as \(ds^2=g_{\mu\nu}dx^{\mu}dx^{\nu}\); explain the shorthand being used here.
Solution: This is nothing more than choosing some chart \(x^{\mu}\) on \(X\) and expanding the metric tensor field \(g\) in the coordinate basis \(\{dx^{\mu}\otimes dx^{\nu}\}\) of type \((0,2)\) tensor fields:
\[g=g_{\mu\nu}dx^{\mu}\otimes dx^{\nu}\]
for some real, symmetric scalar fields \(g_{\mu\nu}:X\to\mathbf R\). One then simply writes \(ds^2:=g\) and omits the tensor product \(\otimes\).
Problem: Let \(x(t)\in X\) be a curve on a Riemannian manifold \((X,g)\). By choosing a chart \(x^{\mu}\) to cover a suitable region of \(X\) spanned by the trajectory \(x(t)\), explain how to compute the length \(L\) of the curve using the Riemannian metric \(g\).
Solution: Heuristically, it is:
\[\int ds=\int\sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}=\int dt\sqrt{g_{\mu\nu}(x(t))\frac{dx^{\mu}}{dt}\frac{dx^{\nu}}{dt}}\]
Problem: Connect this with the action \(S\) of a particle of mass \(m\) moving on a Riemannian manifold \((X,g)\). Hence, derive the geodesic equation and define the Christoffel symbols.
Solution: