3

My question, while related to the posts Ricci flow preserves isometries, The invariance of the Ricci tensor under diffeomorphisms and its non-ellipticity., and Diffeomorphism invariance of the Ricci tensor, is slightly different. Namely, when using Uhlenbeck's trick in the Ricci flow the Riemann tensor evolves by:

$\frac {\partial } {\partial t} R_{abcd} = \iota_a^i \iota_b^j \iota_c^k \iota_d^l\Delta R_{ijkl} + 2(B_{abcd} - B_{abdc} - B_{adbc} + B_{acbd})$

where $\ B_{abcd} = h^{eg}h^{fi}R_{aebf}R_{cgdi}$ and $\iota : (V,h) \rightarrow (TM,g(t)) $ is a 1-parameter family of isometries from the fixed vector space V with metric h to the evolving tangent space TM with metric g(t) such that $ h=\iota^* g(t)$ for all t.

However, according to the diffeomorphism invariance of the curvature $\iota^* Rm[g(t)]=Rm[\iota^* g(t)] $ and $ h=\iota^* g(t) $ and h is fixed. Shouldn't then

$\frac {\partial } {\partial t} R_{abcd} = \frac {\partial } {\partial t} \iota^* Rm[g(t)]=\frac {\partial } {\partial t}Rm[\iota^* g(t)]=\frac {\partial } {\partial t}Rm[h]=0 ?$

Here $\ Rm[g(t)] $ denotes the Riemann tensor of the metric g(t).

Clearly I am not understanding something correctly. What am I missing?

My reference for this material is:

Andrews, Ben, and Christopher Hopper. The Ricci flow in Riemannian geometry: a complete proof of the differentiable 1/4-pinching sphere theorem. springer, 2010.

Jack
  • 31

1 Answers1

0

I think the problem is that you have misunderstood $\iota^*\mbox{Rm}[g]$.

$\iota^*\mbox{Rm}[g]$ is the full pull-back of $\mbox{Rm}[g]$ as a $(4,0)$ tensor, you cannot have $\iota^*\mbox{Rm}[g]=\mbox{Rm}[\iota^*g]$.

Formally if $X,Y,Z,W$ are smooth vector fields over $M$, at the point $p$,

$$\iota^*\mbox{Rm}[g](X,Y,Z,W)(p)=\mbox{Rm}(\iota X,\iota Y,\iota Z,\iota W)(p)=g_p(\mbox{Rm}[g](\iota X,\iota Y)\iota Z,\iota W)$$ $$\mbox{Rm}[\iota^*g](X,Y,Z,W)(p)=g_p(\iota \mbox{Rm}[\iota^*g](X,Y)Z,\iota W)$$

But $\nabla^{\iota^*g}=\nabla^h\neq \iota^*\nabla^g$, and you have changed $X,Y$, therefore $\mbox{Rm}[g](\iota X,\iota Y)\iota Z\neq \iota \mbox{Rm}[h](X,Y)Z$