Are worldlines/geodesics coordinate/frame independent My professor states in the lecture that geodesics are frame independent.
But to me geodesics are just special type of wordlines and worldlines aren't coordinate or frame independent. In the rest frame of the particle the worldline on the Minkowski diagram would be along the time axis but in a moving frame they wouldn't be.
Even a geodesic of a free particle would be a straight line in inertial (cartesian or polar coordinates) but in a rotating or accelerating frame they wouldn't be straight lines.
So why did my professor state that geodesics are frame/coordinate independent.
Will the geodesics/worldlines be (look like ) the same in all (inertial or accelerating or whatever) frames (coordinate system).
Is my understanding wrong. Is what I have said above not correct or does the professor mean something else.
 A: It depends on what you mean here.  I'm fairly certain that what your instructor meant was that whether a worldline is or is not a geodesic is a frame-independent notion.  This is because geodesics $\gamma$ extremize the path length functional:
$$S[\gamma] = \int d\lambda \sqrt{ g(\dot \gamma,\dot \gamma)}$$
Nothing in the definition of $S$ requires a choice of coordinates, so whether $\gamma$ extremizes $S$ or not doesn't depend on your choice of coordinates.
A different way to see this is to note$^\dagger$ that geodesics are solutions to the autoparallel equation
$$\nabla_{\dot \gamma} \dot \gamma = 0$$
which is a tensor equation, so once again coordinate-independent.
Of course, the coordinates of the points along that worldline will clearly depend on which coordinate chart you use.

$^\dagger$This assumes that you have chosen the Levi-Civita connection, which we typically do in GR.
A: The statement you have attributed to your professor is not very rigorous on its own. It is trying to address a common source of confusion amongst students.
Imagine a world without coordinates first. Say you have a ball moving in absence of any forces. GR tells us that it will move along a straight line through spacetime. In the frame attached to the ball there is no motion in space. Time passes uniformly as $\tau$ (proper time). The change in its four velocity ($\mathbf{u}$) with respect to proper time is $\mathbf{0}$ as this ball is a free particle. The following is thus its geodesic equation:
$$
\frac{\mathrm{d}\mathbf{u}}{\mathrm{d}\tau} = \mathbf{0}
$$
Now you sitting somewhere else record this ball's position in space as a function of time. You are therefore introducing a coordinate system. The ball is still moving in a straight line but your coordinate system could be flat or curved depending upon the energy-momentum distribution.
Take a non GR example before we attempt to formalize this further. Imagine an airplane flying the shortest distance from A to B. If you zoom in sufficiently (in other words, just look locally and in formal terms look in the infinitesimal subspace around the plane's position at any point on its trajectory) the velocity vector is parallel to the trajectory. Which implies no change in trajectory. If you now plot this over a globe you get a great circle. If you plot this on Mercator projection you get a strange curve.

Formally, if your coordinate basis is given by $\mathbf{e}_\alpha$ and $\Gamma^\alpha_{\beta\gamma}$ are the Christoffel symbols for your coordinate system then you can simplify the geodesic equation using the chain rule as follows:
$$
\frac{\mathrm{d}\mathbf{u}}{\mathrm{d}\tau} = \frac{\mathrm{d}\mathbf{u}^\alpha \mathbf{e}_\alpha}{\mathrm{d}\tau} = \mathbf{u}^\alpha \frac{\mathrm{d}\mathbf{e}_\alpha}{\mathrm{d}\tau} + \mathbf{e}_\alpha \frac{\mathrm{d}\mathbf{u}^\alpha}{\mathrm{d}\tau} = \mathbf{u}^\alpha \Gamma^\gamma_{\alpha \beta} \mathbf{u}^\beta \mathbf{e}_\gamma + \mathbf{e}_\alpha \frac{\mathrm{d}\mathbf{u}^\alpha}{\mathrm{d}\tau} = \mathbf{u}^\alpha \Gamma^\gamma_{\alpha \beta} \mathbf{u}^\beta \mathbf{e}_\gamma + \mathbf{e}_\gamma\frac{\mathrm{d}\mathbf{u}^\gamma}{\mathrm{d}\tau} = \mathbf{0}
$$
Rearranging you get the geodesic equation in the coordinate form.
$$
\frac{\mathrm{d}\mathbf{u}^\gamma}{\mathrm{d}\tau} = - \Gamma^\gamma_{\alpha \beta} \mathbf{u}^\alpha \mathbf{u}^\beta
$$
Quite clearly, this is a frame dependent equation. But on its own the geodesic is still unchanged. How you plot it has changed.
A: The way physicists generally talk is that things like scalars, vectors, and tensors are classified according to their transformation properties. However, it's understood by the people talking this way that there is an equally valid point of view in which something like a momentum vector is an invariant quantity, but its components have transformation properties that depend on what coordinate system you pick.

Even a geodesic of a free particle would be a straight line in inertial (cartesian or polar coordinates) but in a rotating or accelerating frame they wouldn't be straight lines.

Changing the coordinates doesn't change whether it's curved. A solution to the geodesic equation is still a solution regardless of the coordinates you use.
