[**EDIT:** I rewrote the first couple of paragraphs, because I realized a better way to say what I had in mind.]

There are many ways to define dimension and some of them give the same answer some of them don't.

*Depth* is a sort of dimension. Perhaps not the most obvious, but one that works well in many situation.

In general we count dimension by chains and the main difference between *Krull dimension* and *depth* is about the same as the difference between Weil divisors and Cartier divisors.

For simplicity assume that we are talking about finite dimensional spaces. Infinite dimension can be dealt with by saying that it contains arbitrary dimensional finite dimensional spaces where we may substitute "Krull dimension" or "depth" in place of "dimension".

I usually think of *Krull dimension* as going from small to large: We start with a (closed) point, embed it into a curve, then to a surface until we get to the maximal dimension. However, for comparing to *depth* it is probably better to think of it as going from large to small: Take a(n irreducible) Weil divisor, then a(n irreducible) Weil divisor in that and so on until you get to a point.

In contrast, when we deal with *depth* we take Cartier divisors: We start with the space itself (or an irreducible component), then take a(n irreducible) hypersurface, then the intersection of two hypersurfaces (such that it is a "true" hypersurface in each irreducible component this condition corresponds to the "non-zero divisor" provision)=a codimension $2$ complete intersection, and so on until we reach a zero dimensional set.

So, I would say that the geometric meaning of Cohen-Macaulay is that it is a space where our intuition about these two notions giving the same number is correct. I would also point out that this does not mean that necessarily all Weil divisors are Cartier, just that one cannot get a longer sequence of subsequent Weil divisors than Cartier divisors.

Another, less philosophical explanation is the following:
Cohen-Macaulay means that *depth* = *dimension*. $S_n$ means that this is true up to codimension $n$. Then one may try to give geometric meaning to the $S_n$ property and say that CM means that all of those properties hold.
So,

$S_1$ --- means the existence of non-zero divisors, i.e., that there exists hypersurfaces that are like the ones we imagine.

$S_2$ --- is perhaps the most interesting one, or the one that is the easiest to explain. See this answer to another question where it is explained how it corresponds to the *Hartogs property*, that is, to the condition that functions defined outside a codimension $2$ set can be extended to the entire space.

$S_3$ --- I don't have a similarly nice description of this, but I am sure something could be made up, or some people might even know something nice. One thing is sure. This means that every ("true") hypersurface has the $S_2$ property, which has a geometric meaning as above.

[...]

So, one could say that

$S_n$ means that every ("true") hypersurface has the $S_{n-1}$ property, which we already described.

I realize that this description of $S_n$ may not seem satisfactory, but in practice, this is very useful. I would also add that in moduli theory it is actually important to know that some properties are inherited by hypersurfaces (=fibers of morphisms), so saying that hypersurfaces are $S_2$ *is* actually a good property.

More specifically, for example, the total space of a family of stable (resp. normal, $S_2$) varieties is $S_3$ ("is" as in "has to be"). Then one might (as in Shafarevich's conjecture, see Parshin's Theorem, Arakelov's Theorem, Manin's Theorem, Faltings' Theorem) study *deformations* of these families (say the embedded deformations of a curve in the moduli stack of the corresponding moduli problem). Then the total space of these deformations ought to be $S_4$ on account of their fibers having to be $S_3$ since *their* fibers have to be $S_2$. This actually explains why it is not entirely bogus to say that $S_4$ means that codimension $2$ complete intersections satisfy the Hartogs property.

This actually reminds me another thing that is important about CM. A lot of properties are inherited by *general hypersurfaces*. The CM property is inherited by *all* of them. This makes them perfect for inductive proofs.

One way to see that a surface is CM is that *normal* $\Rightarrow$ *CM*. Of course, the point is that *normal* is equivalent to $R_1$ and $S_2$, so it always implies $S_2$ and if the dimension is at most $2$, then $S_2$ is the same as CM. If you have a non-normal surface, but it is non-normal only because it has normal crossings in codimension one, then it is CM. You may also try to test directly for the Hartogs property mentioned above:

A reduced surface $S$ is Cohen-Macaulay if and only if for any $P\in S$ and any regular function $f$ defined on $U\setminus \{P\}$ for an open set $P\in U\subseteq S$ there exists a regular function $g$ on $U$ such that $g_{|U\setminus \{P\}}=f$.

geometriccharacterization of CM: A surface $S$ is Cohen-Macaulay if and only if for any $P\in S$ and any regular function $f$ defined on $U\setminus \{P\}$ for an open set $P\in U\subseteq S$ there exists a regular function $g$ on $U$ such that $g_{∣U\setminus\{P\}}=f$. $\endgroup$algebra=geometry. In other words, at the end of the day (in algebraic geometry) it is hard to say what ispurely algebraicand what ispurely geometric. But of course, Martin's question is that here is this notion that is clearly an important notion of algebraic geometry and hence both algebraic and geometric. The algebraic interpretation is usually the one we see and the question is whether it can be put in geometric terms, that is, terms that we traditionally associate with geometry. $\endgroup$3more comments