A Riemannian metric on the diffeomorphism group

Question

The (smooth) diffeomorphism group of the unit circle can be defined as
$$\operatorname{Diff}(S^1)=\{\varphi\in C^\infty(S^1,S^1)\,|\,\varphi'>0\}.$$
Identify $S^1$ with the interval $[0,1]$ via $S^1 \cong \mathbb{R}/\mathbb{Z},$ and consider the Lie subgroup $\operatorname{Diff}_0(S^1)$ of $\operatorname{Diff}(S^1)$ defined as
$$\operatorname{Diff}_0(S^1)=\{\varphi\in C^\infty(S^1,S^1)\,|\,\varphi'>0,\,\varphi(0)=0\}.$$
Both of these groups are infinite-dimensional Lie groups, and we can equip $\operatorname{Diff}_0(S^1)$ with the Riemannian metric
$$G_{\varphi}(X,Y)=\int_{S^1}\frac{X'Y'}{\varphi'}d\theta$$
for $\varphi \in \operatorname{Diff}_0(S^1)$ and $X,Y \in T_{\varphi}\operatorname{Diff}_0(S^1),$ where the tangent space consists of smooth vector fields on $S^1$ $\textit{along $\varphi$}.$ Since we're in dimension one, we just identify a vector field with it's component function and treat $X,Y$ as real-valued functions on $S^1=[0,1],$ not vector fields acting on one another.
$\textit{Supposedly,}$ the condition $\varphi(0)=0$ (or one similar in effect) is required to ensure that $G_{\varphi}(X,Y)$ is in fact a Riemannian metric, i.e that it defines an inner product on each of the tangent spaces. But bilinearity, symmetry, and positive-definiteness of $G_{\varphi}(X,Y)$ all seem to obviously be satisfied, independent of the values taken on by $\varphi$. Am I missing something? Do we really need $\varphi(0)=0$ to ensure $G$ is a Riemannian metric?

Answer 1

The theory of infinite dimensional Lie groups is rather involved, so these comments may be somewhat imprecise.

Let $\mathfrak{X}S^1$ denote the space of smooth vector fields on $S^1$ and $\mathfrak{X}_0S^1\subset\mathfrak{X}S^1$ denote the set of vector fields which vanish at $0\in S^1$. $\mathfrak{X}S^1$ is the Lie algbebra of $\operatorname{Diff}(S^1)$, while $\mathfrak{X}_0S^1$ is the Lie algbebra of $\operatorname{Diff}_0(S^1)$ (since the flow of a vector field $X$ fixes $0$ iff $X(0)=0$). The inner product you propose is in fact not positive definite on $\mathfrak{X}S^1$, since $G_{\operatorname{id}}(X,X)=0$ for any constant vector field $X$. It is, however, positive definite on $\mathfrak{X}_0S^1$.