Existence of Orthonormal Basis of a Metric in a Manifold

Definition: A metric $g$ on a manifold $M$ is a tensor field of type $(0,2)$ such that

(1) it is symmetric, i.e. $g(v,w)=g(w,v)$ for any $w,v \in V_p, p\in M$, and

(2) it is non-degenerate, i.e. if $v_1 \in V_p$ such that $g(v_1,v)=0$ for any $v \in V_p$, then $v=0$.

($V_p$ denotes the tangent space at $p$)

Theorem: Given a metric $g$, there exists some orthonormal basis $v_1,...,v_n$ for $T_p$ for each $p \in M$, i.e. $g(v_\mu,v_\nu)=\pm \delta_\mu^\nu$. Moreover, if $S,S'$ are two orthonormal basis, then

$$\#\{s \in S: g(s,s)=1\}=\#\{s' \in S': g(s',s')=1\}$$

and

$$\#\{s \in S: g(s,s)=-1\}=\#\{s' \in S': g(s',s')=-1\}$$

For the first part, I am wondering there is anything like the Gram-Schmidt process which can be applied to this case, but since the metric may not be positive definite, I can't see how I can do so.

For the second part, I do not have any idea how to approach it.

Thank you for providing me suggestions.

• Sylvester's theorem – Tom Chalmer Aug 5 '18 at 23:25

Suppose we have a Semi-Riemannian manifold $(M^n,g)$ with metric signature $(n-k,k)$. By definition, each $p \in M$ the map $g_p : T_pM \times T_pM \to \Bbb{R}$ is a non-degenerate, symmetric, bilinear form. We can find an orthonormal basis for $T_pM$ for any $p \in M$ using induction as shown in this lemma (lemma 24 of Barrett O'neill's Semi-Riemannian Geometry, p.50).

For the second question, it's enough to do that in the level of vector space. Suppose $V$ is a $n$-dimensional real vector space endowed with a non-degenerate, symmetric bilinear form $g : V \times V \to \Bbb{R}$. Let $\{e_1,\dots,e_n\}$ is an arbitrary orthonormal basis for $V$.

Let $k \leq n$ be the number of negative values in $\{g(e_i,e_i)=\pm1 : i=1,\dots ,n\}$. The case $k=0$ is trivial. If $k>0$, then $V$ will have a subspaces which $g$ is negative definite, e.g. the span of one of the basis elements. Let $W$ be the subspace of maximal dimension on which $g$ is negative definite. If we can show that $k=\text{dim }W$ then we're done since this number $k$ is independent of basis.

By rearranging the basis $\{e_1,\dots,e_k,e_{k+1},\dots,e_n\}$, we have $g(e_i,e_i)=-1$ for $1\leq i \leq k$ and $g(e_j,e_j)=1$ for $k+1\leq j \leq n$. Since $g$ negatives definite on $X = \text{span} (e_1,\dots,e_k)$ then $k=\text{dim }X \leq \text{dim }W$. To show $k \geq \text{dim }W$, define a map $T : W \to X$ as follow; for any $w = \sum_{i=1}^n w^i e_i \in W$ $$T(w) = \sum_{i=1}^k w^ie_i.$$ You can check directly that $T$ is injective and therefore $\text{dim }W = 0 + \text{dim Im }T \leq \text{dim }X=k$. Therefore the number $k$ is fixed by $g$, independent of basis.

• How do you define $|v|$ for a tangent vector? I suppose a tangent vector $v$ is a function. – Jerry Aug 5 '18 at 17:02
• @Jerry Ah. I think this is not gonna work, since $|\cdot|$ may be zero on some places on $U$. It is defined as $|\cdot| = \sqrt{|g(\cdot,\cdot)|}$. I will delete this answer. – Sou Aug 5 '18 at 17:06
• Mind leaving your answer to the second part here? That means only delete the first part but not the second. – Jerry Aug 5 '18 at 17:12
• I just leave it that way. – Sou Aug 5 '18 at 17:18
• @Jerry I have corrected my answer. I think this okay now. I'm sorry i misread your question. If you mean orthonormal basis just for a tangent space, then it's done in lemma 24 of barrett o'neill's (as linked above). My answer is kind of overkill since it's about construction of local orthonormal frame. – Sou Aug 5 '18 at 22:38

For any symmetric bilinear form (not necessarily nondegenerate) $$\langle\_,\_\rangle$$ on a finite-dimensional vector space $$V$$ over $$\mathbb{R}$$, I claim that there exists a basis of $$V$$, called a good basis, consisting of $$u_1,u_2,\ldots,u_p,v_1,v_2,\ldots,v_q,w_1,w_2,\ldots,w_r\in V$$ such that

• $$p+q+r=\dim_\mathbb{R}(V)$$
• $$\langle u_i,u_j\rangle =+\delta_{i,j}$$ for $$i,j=1,2,\ldots,p$$,
• $$\langle v_i,v_j\rangle = -\delta_{i,j}$$ for $$i,j=1,2,\ldots,q$$,
• $$\langle u_i,v_j\rangle=0$$ for $$i=1,2,\ldots,p$$ and $$j=1,2,\ldots,q$$, and
• $$\langle x,w_k\rangle=0$$ for all $$x\in V$$ and $$k=1,2,\ldots,r$$.

Here, $$\delta$$ is the Kronecker delta. Thus, in the basis above, the bilinear form is represented by the matrix $$J_{p,q,r}:=\begin{bmatrix} +I_{p\times p}&0_{p\times q}&0_{p\times r}\\ 0_{q\times p}&-I_{q\times q}&0_{q\times r}\\ 0_{r\times p}&0_{r\times q}&0_{r\times r}\end{bmatrix}\,,$$ where $$I_{k\times k}$$ is the $$k$$-by-$$k$$ identity matrix and $$0_{\alpha\times \beta}$$ is the $$\alpha$$-by-$$\beta$$ zero matrix. For each $$x\in V$$, we write $$\|x\|$$ for $$\sqrt{\big|\langle x,x\rangle\big|}$$, and write $$\sigma(x)\in\{-1,0,+1\}$$ for the sign of $$\langle x,x\rangle$$.

First, let $$W$$ be the kernel of the bilinear form. That is, $$W$$ consists of all vectors $$z\in V$$ for which $$\langle x,z\rangle=0$$ for all $$x\in V$$. Let $$r$$ denote the dimension of $$W$$ over $$\mathbb{R}$$. We can take $$\left\{w_1,w_2,\ldots,w_r\right\}$$ to be any basis of $$W$$. (This also shows that $$r$$ is independent of the choice of good basis of $$V$$, as it must be the $$\mathbb{R}$$-dimension of $$W$$, a fixed subspace of $$V$$.) We can from now on assume that $$W=0$$ (that is, the bilinear form $$\langle\_,\_\rangle$$ is nondegenerate). Otherwise, we study the vector space $$V/W$$ with the bilinear form $$\langle\!\langle\_,\_\rangle\!\rangle$$ defined by $$\langle\!\langle x+W,y+W\rangle\!\rangle:=\langle x,y\rangle\text{ for all }x,y\in V\,.$$

Fix a basis $$\left\{z_1,z_2,\ldots,z_n\right\}$$ of $$V$$. Write $$[n]:=\{1,2,\ldots,n\}$$. We shall perform the following orthonormalization procedure. First, we look at $$\|z_1\|$$. If $$\|z_1\|=0$$, then note that, for some index $$j\in[n]\setminus\{1\}$$, $$\langle z_1,z_j\rangle \neq 0$$, (as the bilinear form is nondegenerate), and so we can replace $$z_1$$ by $$z_1+tz_j$$ where $$t\in\mathbb{R}\setminus\{0\}$$ is so chosen that $$2\,\langle z_1,z_j\rangle +t\langle z_j,z_j\rangle \neq 0$$. Hence, we may always assume that $$\|z_1\|\neq 0$$. Dividing $$z_1$$ by $$\|z_1\|$$, we may also assume that $$\|z_1\|=1$$. Now, we replace $$z_j$$ for $$j=2,3,\ldots,n$$ by $$z_j-\langle z_j,z_1\rangle z_1\,.$$ By doing so, we may assume that each $$z_j$$ is orthogonal to $$z_1$$ already.

Let $$V_1$$ denote the orthogonal complement of $$z_1$$. That is, $$V_1$$ consists of all vectors $$x$$ in $$V$$ such that $$\langle x,z_1\rangle=0$$. Clearly, $$V_1$$ is an $$(n-1)$$-dimensional $$\mathbb{R}$$-subspace of $$V$$. By the paragraph above, $$V_1$$ is the $$\mathbb{R}$$-span of $$z_2,z_3,\ldots,z_n$$, and $$V_1$$ inherits the bilinear form $$\langle\_,\_\rangle_1$$ from $$V$$ (by simply restricting $$\langle \_,\_\rangle$$ onto $$V_1\times V_1$$). We can then repeat the paragraph above for $$V_1$$, noting that $$\langle\_,\_\rangle_1$$ is nondegenerate. Hence, by induction, you may assume that the vectors $$z_2,z_3,\ldots,z_n$$ are orthogonal, and each $$\|z_j\|$$ is equal to $$1$$ for $$j=2,3,\ldots,n$$.

Then, we let $$u_1,u_2,\ldots,u_p$$ to be the vectors $$z_j$$ with $$\sigma(z_j)=+1$$ ($$j\in[n]$$). Likewise, $$v_1,v_2,\ldots,v_q$$ are the vectors $$z_j$$ with $$\sigma(z_j)=-1$$ ($$j\in[n]$$). Note that $$p$$ and $$q$$ are also independent of the choice of good bases. However, it is easier to show that, if $$S$$ is an $$n$$-by-$$n$$ real symmetric and $$A$$ is an $$n$$-by-$$n$$ invertible real matrix, then $$A^\top\,S\,A$$ and $$S$$ have the same number of positive eigenvalues (with multiplicities), and the same number of negative eigenvalues (with multiplicities). The triple $$(p,q,r)$$ is called the signature of a symmetric bilinear form.