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.

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    $\begingroup$ What is a "conformal manifold"? $\endgroup$ Apr 24 at 8:49
  • $\begingroup$ 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. $\endgroup$
    – Boka Peer
    Apr 24 at 15:50

1 Answer 1


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.

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    $\begingroup$ 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. $\endgroup$
    – Boka Peer
    Apr 24 at 15:36
  • $\begingroup$ 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. $\endgroup$
    – Boka Peer
    Apr 25 at 3:41
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    $\begingroup$ Your explanation for the degeneracy of the metric is correct. $\endgroup$ Apr 25 at 11:01

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