If the notes seem a bit thin for this lecture, it's because I went over some stuff that can be found in the notes from Lecture 23 (May 20) that I didn't actually do in that class.
So far, it seems like every question we ask about Turing machines is undecidable. As it turns out, this is not just an educated hunch, but a theorem. First, we'll try to formalize the notion of a "property".
A property $P$ of a Turing machine is a semantic property if it depends on the language recognized by $M$ and not the syntactic structure of $M$. In other words, if $L(M_1) = L(M_2)$, then $M_1$ and $M_2$ have the same semantic properties.
A property $P$ can be expressed as a language consisting of exactly the encodings $\langle M \rangle$, where $M$ has property $P$. A semantic property $P$ is said to be non-trivial if there exists a TM $M_1$ such that $\langle M_1 \rangle \in P$ and a TM $M_2$ such that $\langle M_2 \rangle \not\in P$.
Then we have the following theorem.
All non-trivial semantic properties of Turing machines are undecidable.
In other words, for any given Turing machine $M$, we can't decide any properties about $L(M)$ except for properties that are true for either exactly all or exactly none of the languages recognized by Turing machines. The proof is surprisingly simple, since it follows the basic template that we've been using so far.
To show this, we assume that $P$ is a decidable language that satisfies some property and we let $R_P$ be a TM that decides $P$. We will reduce TM membership to property testing.
Let $T_\emptyset$ be a TM that always rejects, so $L(T_\emptyset) = \emptyset$. Without loss of generality, we assume that $\langle T_\emptyset \rangle \not\in P$, since otherwise, we can just use $\overline P$ instead. Since $P$ is non-trivial, there exists a TM $T$ with $\langle T \rangle \in P$.
We construct a Turing machine that decides $A_{TM}$ by being able to distinguish between $T_\emptyset$ and $T$.
So $M_w$ simulates $T$ if $M$ accepts $w$. Then $L(M_w) = L(T)$ if $M$ accepts $w$ and $L(M_w) = \emptyset$ otherwise. Thus, $\langle M_w \rangle \in P$ iff $M$ accepts $w$.
Rice's theorem was proved by Rice in 1951 (the result appeared in a journal later in 1953). It is interesting to note that a similar result was discussed by Turing in his original 1936 paper about undecidability.
There is a similar result characterizing undecidable properties of languages for subclasses of decidable languages (i.e. context-free languages, context-sensitive languages, etc.) called Greibach's theorem. As you might expect, this was shown by Sheila Greibach in 1963 (you may remember Greibach being mentioned when were talking about normal forms for grammars).
We won't prove Greibach's theorem, but we will show that some problems about context-free languages are undecidable. The proofs for these kinds of problems are going to be much trickier, since we don't have the benefit of conjuring up Turing machines. However, we can do something that may be surprising, which is make use of computations of Turing machines.
Let $M$ be a Turing machine and $w$ and input string. An accepting computation path for $M$ on $w$ is a sequence of configurations $C_1, \dots, C_\ell$, where $C_1$ is the start configuration of $M$ on $w$ and $C_\ell$ is an accepting configuration of $M$ on $w$ and each $C_i$ follows directly from $C_{i-1}$ (that is, $C_{i-1} \vdash C_i$) as defined by the transition function of $M$. We define a rejecting computation path similarly.
This gives us a very strict format for the computation of a word $w$ when run on a Turing machine $M$. We will show that we can compute all of the strings that aren't accepting computations.