I am stuck at Problem 1.4(a) from Riemannain Geometry, written by Do Carmo.

A function $g:\mathbb R\to\mathbb R$ given by $g(t)=yt+x$, $t$,$x$,$y\in\mathbb R$, $y>0$, is called a proper affine function. The subset of all such function with respect to the usual composition law forms a Lie group $G$. As a differentiable manifold $G$ is simply the upper half-plane $\{(x,y)\in\mathbb R^2|y>0\}$ with the differentiable structure induced from $\mathbb R^2$. Prove that:

The left-invariant Riemannian metric of $G$ which at the neutral element $e=(0,1)$ coincides with the Euclidean metric ($g_{11}=g_{22}=1,g_{12}=0$) is given by $g_{11}=g_{22}=\frac{1}{y^2},g_{12}=0$.

I calculate the left-invariant Riemannian metric from its definition $$g_{11}=\left\langle\frac{\partial}{\partial x}|_{(x,y)},\frac{\partial}{\partial x}|_{(x,y)}\right\rangle_{(x,y)}=\left\langle(L_{(x,y)}^{-1})_*(\frac{\partial}{\partial x}|_{(x,y)}),(L_{(x,y)}^{-1})_*(\frac{\partial}{\partial x}|_{(x,y)})\right\rangle_e$$

Now $G$ is a half-plane with the composition law: $(x_1,y_1)(x_2,y_2)=(x_2+x_1y_2,y_1y_2)$. So $$L_{(x,y)}^{-1}(a,b)=\left(a-\frac{x}{y}b,\frac{1}{y}b\right)$$

Then $g_{11}=\left\langle\frac{\partial}{\partial x}|_e,\frac{\partial}{\partial x}|_e\right\rangle_e=1$

This isn't equal to $\frac{1}{y^2}$. Where do I make mistakes? Any advice is helpful. Thank you.


1 Answer 1


$$ G=\{ g_{xy}:=(x,y)| y>0 \} $$ Here multiplication is give by $$ g_{xy}\circ g_{pq} (t)= g_{xy} ( qt+p)=y(qt+p)+ x =yq t+(py+x) $$

Since $yq>0$ so $(x,y)\cdot (p,q)=(py+x,yq)$

That is $$ \frac{d}{dt}(p,q)=(v,w)\Rightarrow \frac{d}{dt}(x,y)\cdot (p,q)=y(v,w) $$

If $h$ is left invariant metric and if $\frac{d}{dt} (p_1,q_1)=(v_1,w_1)$ then $$h\bigg( \frac{d}{dt} (x,y)\cdot (p,q)(t), \frac{d}{dt} (x,y)\cdot (p_1,q_1)(t)\bigg)=h\bigg( \frac{d}{dt} (p,q)(t), \frac{d}{dt} (p_1,q_1)(t)\bigg) $$ so that $$ y^2 h\bigg((v,w),(v_1,w_1)\bigg) (p(0)y+x,yq(0))= h\bigg((v,w),(v_1,w_1)\bigg)(p(0),q(0)) $$

  • 1
    $\begingroup$ why make sense think in $\frac{d}{dt} (p,q)$? I know that $(p,q)$ can be written as $g(t) = qt + p$, but how this relation allow us think in $\frac{d}{dt} (p,q)$? $\endgroup$
    – George
    Jan 8, 2018 at 12:38

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .