Why is $g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}$ a constant for geodesics in GR? In Sean Carroll's spacetime and geometry chapter 5 Carroll states the following

In addition we always have another constant of the motion for geodesics: the geodesic equation (together with metric compatibility) implies that the quantity$$\epsilon=-g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}\tag{5.55}$$ is constant along the path. (For any trajectory we can choose the parameter $\lambda$ such that $\epsilon$ is a constnat; we are simply noting that this is compatible with affine parameterization along a geodesic.)

I feel like I understand this now but my understanding of formal tensors is still a bit shaky so could you tell if my reasoning is valid and correct any misunderstandings?

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*From the metric components $g_{\mu\nu}$ we can construct the metric tensor given by $g(x)=g_{\mu\nu}(x)dx^\mu\otimes dx^\nu$. Since it is a tensor it is reparametrization invariant but can still vary over space.

*Metric compatibility tells us that actually it doesn't vary over space. Since $\nabla_\alpha g_{\mu\nu}=0$ the metric can be parallel transported to any point in space so it is constant.

*If $x(\lambda)$ is a geodesic then from the geodesic equation we have $\nabla_{\dot x}\dot x=0$. This means the direction of the tangent vector $\frac{d}{d\lambda}$ is conserved. By a suitable (reparametrization) of $\lambda$ we get that $\frac{d}{d\lambda}$ is conserved.

*Since $g_{\mu\nu}dx^\mu\otimes dx^\nu$ and $\frac{d}{d\lambda}$ are both conserved we have that
$$g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}=g\left(\frac{d}{d\lambda}\otimes \frac{d}{d\lambda}\right)=\left(g_{\mu\nu}dx^\mu\otimes dx^\nu\right)\left(\frac{d}{d\lambda}\otimes \frac{d}{d\lambda}\right)$$
is also conserved.

 A: You don't need any knowledge of tensor calculus to understand this.
What Does   $g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^
\nu}{d\lambda}$ Represent?
For a curve $x^\mu(\lambda)$ in a manifold, the quantity $$g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^
\nu}{d\lambda}$$  is the square of the length of the tangent vector $\textbf{t}$ to the curve $x^\mu(\lambda)$ at any point $P$.
To see this, note that at any point $P$ on the curve, the tangent vector $\textbf{t}$  is defined as
$$\textbf{t}=\frac{d\textbf{s}}{d\lambda},$$
where $d\textbf{s}$ is the infinitesimal separation vector between point $P$ and a nearby point $Q$ on the curve corresponding to the parameter value $\lambda+d\lambda$.
In a given coordinate system with basis vectors $\textbf{e}_\mu$, we can write $d\textbf{s}=\textbf{e}_\mu dx^\mu$ so that the tangent vector is now
$$\textbf{t}=\frac{dx^\mu}{d\lambda}\textbf{e}_\mu.$$
(Note: More explicitly, $dx^\mu \equiv dx^\mu(\lambda)$.)
The square of the length of the tangent vector $\textbf{t}$ is then
$$|\textbf{t}|^2=g_{\mu\nu}t^\mu t^\nu=g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^\nu}{d\lambda}.$$
So the quantity $g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^
\nu}{d\lambda}$ being constant throughout the curve $x^\mu(\lambda)$ means that the tangent vector $\textbf{t}$ has a constant length throughout the curve.
Parameterising A Curve
For any curve $x^\mu(\lambda)$, we can paramaterise the curve such that the length of the tangent vector is constant. Note that if we choose the parameter $\lambda$ to be
$$\lambda=as+b,$$
where $s$ is the distance measured along the curve and $a$, $b$ are constants, the length of the tangent vector will be constant. This can be shown through
$$|\textbf{t}|=\frac{d|\textbf{s}|}{d\lambda}=\frac{ds}{d\lambda}={1\over a},$$
where $1\over a$ is a constant.
Summary: For any curve $x^\mu(\lambda)$, we can always parameterise it such that the length of the tangent vector is constant throughout the curve. The length of the tangent vector is given by $|\textbf{t}|=(g_{\mu\nu}\frac{dx^\mu}{d\lambda}\frac{dx^
\nu}{d\lambda})^{1\over2}$.
References:

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*Hobson, Efstathiou & Lasenby General Relativity: An Introduction for Physicists pg. 75

A: This is simply a choice of parametrization of the geodesic $x^\mu(\lambda)$.  If we were in a Euclidean-signature manifold, $\lambda$ would be proportional to the arc length along the curve. Here, for a timelike geodesic, it would be proportional to the  proper time along the curve.
