TITLE: Are there Type III codes with small but nonzero "index"?
QUESTION [8 upvotes]: Recall that a Type III code of rank $r$ is a linear subspace $C \subset \mathbb F_3^r$ which is self-dual for the standard inner product. (These occur only when $r$ is divisible by $4$.) Elements of $C$ are called code words. The Hamming weight of a code word is its number of non-zero entries. I will call a code word maximal if its Hamming weight is $r$. (Maximal codewords clearly only occur when $r$ is divisible by $12$.)
The set of maximal codewords can be partitioned into two subsets as follows: two maximal words $w_1,w_2$ are in the same subset if $w_1-w_2$ has even Hamming weight, and are in opposite subsets if $w_1-w_2$ has odd Hamming weight. Call these subsets $M_+$ and $M_-$.
I suggest the following definition:
Definition: The index of $C$ is $|M_+| - |M_-|$. Of course, this is only defined up to sign. I could take its absolute value if I cared. The name is because this is related to the "supersymmetric index" of a certain supersymmetric field theory.
The discussion in the comments of this question implies the following when $r$ is divisible by $24$:
Proposition: The index of any Type III code of rank $r=24k$ is divisible by $24$.
Conjecture: This is true also when $r \equiv 12 \mod 24$. (It is trivially true for $r \equiv \pm 4 \mod 12$, as then $M_\pm$ are empty.) 
Moreover, I suspect that both $|M_+|$ and $|M_-|$ are necessarily divisible by $24$.
Example: The Ternary Golay code, with rank $12$ has index $24$. It is the unique rank-$12$ code with non-zero index.
Example: The complete classification of Type III codes of rank $24$ is known. Assuming I read it correctly, there are precisely five codes of rank $24$ and index $24$.

Question: In higher rank, do there exist Type III codes with index exactly $24$? For example, what about rank $36$?

At best, there would be some general algorithm that produces a code with index $24$ for each rank $r = 12k$.
By the way, I know how to prove:
Proposition: If the code $C$ contains words of Hamming weight $3$, then its index vanishes.
So if you are looking for such a code, you know not to look at such codes.
It's pretty easy to show:
Lemma: The index multiplies when you take the direct sum of codes.
Since $24^2$ is pretty big, you probably will want to work with indecomposable codes.

REPLY [4 votes]: Index $24$ isn't hard for length $36$.
For example, the Type III code with generator matrix
+ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 + - + 0 - - + 0 + + - + + - 0 -
0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 - 0 - + - - 0 0 - 0 0 0 0 - + - - -
0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 - + 0 + - - - + + - + 0 - 0 - +
0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 + 0 + - + 0 0 + 0 + 0 - - + -
0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 0 - + 0 0 - 0 - + 0 + 0 - - 0 + 0 + -
0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 0 + - 0 - - 0 + + 0 + 0 + 0 - 0 - + 0
0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 - + 0 0 - + 0 - - 0 0 - + - 0 0 - +
0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 0 0 + + - - - - - + 0 + - - - 0 + - 0
0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 0 + + - - + + 0 + - 0 + 0 0 + + + + -
0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 0 - + + + 0 0 - + + - + - - 0 + 0 - -
0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 0 + + 0 0 0 - + + + + + 0 - 0 - 0 0 +
0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 + 0 0 0 - - + 0 0 - 0 + - - + 0 - +
0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 0 0 + 0 - 0 0 - + - - 0 - + + - 0 0 +
0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 0 + 0 - + - - - - 0 0 - 0 - - - - - -
0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 0 - 0 - - - 0 + + 0 + - 0 - 0 - + 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 0 + - - + 0 0 - 0 - + - + - - - + + 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 - 0 + - 0 - 0 + 0 - - 0 + 0 0 - - -
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 + 0 + + + + - + 0 - - 0 + 0 + - - - +

has index $24$.  (The total number of maximal codewords is $520$; of
$70$ "random" Type III codes I tried, $26$ had index $24$ -- which indeed 
was the most common index -- and all had index divisible by $24$, 
while the number of maximal codewords was always a multiple of $8$ 
but not necessarily $24$.)