# Definition of the conformal metric

On page 16 of this lecture notes and in this lecture, at some point (somewhere at 1:37:45), Rod Gover defined the $$\text{conformal metric}$$ $$\mathbb{g}$$ on a conformal manifold $$(M, [g])$$. He mentioned that this metric is canonical. Does this mean it is an invariant of the conformal manifold $$(M, [g])$$? Also, he mentioned that $$\text{conformal metric}$$ $$\mathbb{g}$$ gives a $$1-1$$ correspondence between the positive section of conformal $$1$$ densities and the metrics of $$[g].$$ Since I don't understand the definition, I cannot see this correspondence.

Could you please elaborate on this definition of $$\text{conformal metric}$$ $$\mathbb{g}$$ on a conformal manifold $$(M, [g])$$? Note that on page $$12$$ of this, one can find the definition, but to me, the definition is not well stated.

I would appreciate any detailed answer from the scratch. Note that I am comfortable with the associated bundle associated with a given principal bundle and conformal densities of weight $$w.$$ We can visualize/interpret a conformal manifold $$(M, [g])$$ as a ray bundle, an $$\mathbb{R_{>0}}-$$ principal bundle and its sections correspond to the metrics $$g \in [g]$$. I would appreciate any help you could provide. Thanks so much.

NB:I feel like this problem is too basic for this site. If you find an answer on MathStackExchange, please transfer this post to MathStackExchange.

• What is a "conformal manifold"? Apr 24 at 8:49
• Sorry, I forgot to add the definition. You can see that Jeffrey Case has written the definition of conformal manifold in his first sentence of the answer below. Apr 24 at 15:50

Let $$(M,[g])$$ be a conformal manifold; i.e. $$(M,g)$$ is a Riemannian manifold and $$[g] = \{ u^2g \mathrel{}:\mathrel{} u \in C^\infty(M), u>0 \}$$ is the set of Riemannian metrics conformal to $$g$$. It is clear that $$[g] = [u^2g]$$ for all positive $$u \in C^\infty(M)$$. Let me now change notation and write $$c:=[g]$$, to emphasize that there is no canonical choice of representative of the given conformal class.

A conformal manifold $$(M,c)$$ determines a $$\mathbb{R}_+$$-subbundle $$\mathcal{Q} \subset S^2T^\ast M$$ by $$\mathcal{Q} = \{ (x,g_x) \mathrel{}:\mathrel{} x \in M, g \in c \} .$$ Define the action of $$\mathbb{R}_+$$ on $$\mathcal{Q}$$ by $$\delta_s(x,g_x) := (x,s^2g_x)$$ for $$s>0$$. Let $$\pi \colon \mathcal{Q} \to M$$ be the projection $$\pi(x,g_x)=x$$ inherited from $$S^2T^\ast M$$. Then $$\mathcal{Q}$$ is a principal $$\mathbb{R}_+$$-bundle.

The canonical metric is the degenerate metric $$\mathbf{g}$$ on $$\mathcal{Q}$$ defined as follows: Let $$X,Y \in T_{(x,g_x)}\mathcal{Q}$$ be tangent vectors at a point $$(x,g_x)$$ in $$\mathcal{Q}$$. Then $$\pi_\ast X, \pi_\ast Y \in T_xM$$, and so we may define $$\mathbf{g}(X,Y) := g_x(\pi_\ast X, \pi_\ast Y) .$$ Since this metric depends only on $$\mathcal{Q}$$, it is an invariant of $$(M,c)$$.

Next observe that if $$X \in T_{(x,g_x)}\mathcal{Q}$$ and $$s>0$$, then $$(\delta_s)_\ast X \in T_{(x,s^2g_x)}\mathcal{Q}$$ and $$\pi_\ast (\delta_s)_\ast X = (\pi \circ \delta_s)_\ast X = \pi_\ast X ,$$ where the last equality uses the observation that $$\pi \circ \delta_s = \pi$$. It readily follows that if $$X,Y \in T_{(x,g_x)}\mathcal{Q}$$, then $$(\delta_s^\ast \mathbf{g})(X,Y) = \mathbf{g}((\delta_s)_\ast X, (\delta_s)_\ast Y) = s^2g_x(\pi_\ast X,\pi_\ast Y) = s^2\mathbf{g}(X,Y) .$$ That is, $$\mathbf{g}$$ is homogeneous of degree $$2$$ with respect to the dilations $$\delta_s$$.

Finally, a positive conformal density of weight $$1$$ may be regarded a positive function $$\sigma \in C^\infty(\mathcal{Q})$$ such that $$\delta_s\sigma = s\sigma$$ for all $$s \in \mathbb{R}_+$$. It immediately follows that $$\delta_s^\ast (\sigma^{-2}\mathbf{g}) = \sigma^{-2}\mathbf{g}$$. Conversely, if $$h \in c$$ is a choice of conformal metric, then $$\underline{h}(X,Y) := h(\pi_\ast X, \pi_\ast Y),$$ $$X,Y \in T_{(x,g_x)}\mathcal{Q}$$, is such that $$\delta_s^\ast\underline{h}=\underline{h}$$. It follows that there is some positive $$\sigma \in \Gamma(\mathcal{E}[1])$$ such that $$\underline{h} = \sigma^{-2}\mathbf{g}$$. This is the correspondence between choices of metric in $$c$$ and positive sections of $$\mathcal{E}[1]$$ indicated by Curry and Gover.

• What a surprise! When I was asking this question, I told my friend that it would be great if Jeffrey Case writes an answer. I just like the way you answer questions on this site. Thanks again, Jeffrey Case. Apr 24 at 15:36
• The degeneracy of the metric $\mathcal{g}$ is due to the fact that for any two non-zero "vertical" vectors X and Y at $(x, g_x),$ we $\mathcal{g}(X, Y)= 0,$ right? Thank you very much for this beautiful answer. This answer answers all the questions related to conformal metric $\mathcal{g}$ that I had. Apr 25 at 3:41
• Your explanation for the degeneracy of the metric is correct. Apr 25 at 11:01