(Lower) homotopy groups from triangulations Both cohomology and homotopy groups capture global topological information of a manifold $X$. It is interesting to ask if they can be computed from local data. A triangulation $T$ is a natural presentation of a manifold.
The cohomology by definition can be computed from $T$. However, the problem seems much harder for homotopy groups. In fact, humans struggle even for the simplest case $X=S^n$, at least for $\pi_{k>n}(X)$.
Hope is not lost, as we know the answer to $k \leq n$ for spheres. Therefore my question:
Question: Given a triangulated (compact) manifold $X$ of dimension $n$, is there an algorithm that computes $\pi_{k \leq n}(X)$ (up to isomorphism, or in the form of generators and relations)?
 A: Being a manifold or the dimension restriction $k\leq n$ doesn't matter, the following applies to finite simplicial complexes in general:
As others have explained, if the fundamental group is not finite, the higher homotopy groups might be infinitely generated, so in that case even the format of the answer is unclear. Also, it is known that deciding whether a group given by a finite presentation (the simplicial complex gives you such a presentation for $\pi_1$) is actually finite is not possible algorithmically (the halting problem can be reduced to this problem, so if this is problem would admit an algorithmic solution, the halting problem would, too). This problem arises even if you restrict attention to manifolds: Any finite presentation of a group can be turned into an explicit manifold of dimension $4$ or greater via handlebodies.
However, there are algorithms that will produce the list of elements in finite time if $\pi_1$ happens to be finite (and simply never terminate otherwise). From this, it is possible to produce a finite simplicial complex describing the universal cover. For simply-connected finite simplicial complexes, the problem of computing any given homotopy group is known to be be algorithmic, in fact, there exist actual implementations. (I do think they scale pretty badly with degree though).
EDIT: For the $\pi_1$ statement, let me give you an easy but horribly inefficient algorithm. For a finitely presented group $G$, and a finite group $H$ given by a multiplication table, a map $G\to H$ can be specified by finite data: Pick an image for each generator, such that the relations are satisfied. Conversely, a map $H\to G$ can be described by choosing for each element of $H$ a word in the generators of $G$, and for each pair of elements $h_1,h_2\in H$, a finite sequence of rewritings using the relations of $G$, identifying the word associated to $h_1h_2$ with the product of words associated to $h_1,h_2$. For two such maps, a proof witnessing that the composite $G\to H\to G$ is the identity can similarly be specified in a finite amount of data: For each generator, a finite sequence of rewrites identifying it with its image. Similarly for the other composite. So a proof identifying a finitely presented group with a concrete finite group $H$ can be given as a finite bundle of data, now just enumerate all such bundles by size until you found one.
A: Collins and Miller proved that it is algorithmically undecidable whether or not $\pi_2(X)$ is trivial, for $X$ a finite 2-complex. As Achim Krause points out, in general $\pi_2(X)$ is only a module over $\pi_1(X)$, and so it's not clear what kind of a description of it you might want. But the Collins--Miller result shows that any such description that works in general won't be good enough to determine whether or not $\pi_2(X)$ is trivial (much as knowing the generators and relations for $\pi_1(X)$ doesn't enable you to decide whether or not $\pi_1(X)$ is trivial).
You might also like to bear the Whitehead asphericity conjecture in mind.

Whitehead's conjecture: If a 2-complex $X$ is aspherical and $Y$ is a subcomplex of $X$, then $Y$ is also aspherical.

This has been open for 81 years, so is often reckoned to be one of the hardest questions in topology.
For finite $X$, the question can be rephrased as:

If $X$ is obtained by attaching a 2-cell to $Y$ and $\pi_2(Y)$ is non-trivial, must $\pi_2(X)$ be non-trivial?

As this makes clear, we don't really understand the effect that attaching a 2-cell can have on $\pi_2$.
