I am learning topological data analysis on my own. I am currently basically watching This course. But there this thing in the course note that I didn't understand.
So for this persistence module:
$$ \mathbb{K}^1 \stackrel{\quad\left(\begin{array}{lll}1 \\ 0 \\ 0\end{array}\right) \quad}{\longrightarrow} \mathbb{K}^3 \stackrel{\quad\left(\begin{array}{lll}1 & 1 & 1\end{array}\right) \quad}{\longrightarrow} \mathbb{K}^1 \longrightarrow \mathbb{K}^1 ... $$
A theorem tells me that I can decompose this as direct sum of interval modules in an unique way. But I don't know how this it possible for this example nor how this is unique. In the course note, it is decomposed as intervals [1, inf], [2, 3], [2, 3]. But I think the direct sum of these would be
$$ \mathbb{K}^1 \stackrel{\quad\left(\begin{array}{lll} 1 \\ 0 \\ 0\end{array}\right) \quad}{\longrightarrow} \mathbb{K}^3 \stackrel{\mathrm{Id}}{\longrightarrow} \mathbb{K}^3 $$
Moreover, if [1, inf], [2, 3], [2, 3] works, why [1, 3], [2, 3], [2, inf] doesn't work? So how come the uniqueness?
I have also an additional question, after decomposing, the intervals that I get are the barcodes, right?
Please help me. Thanks!
Theorem 1. A persistence module $\mathbb{V}$ can be decomposed as a direct sum of interval module, written as: $$ \mathbb{V} \simeq \bigoplus \mathbb{I}_{\left[b_j, d_j\right]} $$ in the following case (sufficient, not necessary):
- If $T$ is finite. [Gabriel, 72].
- When all the vector spaces $V_t$ are finite-dimensional, [Crowley-Boevey, 2012]. Furthermore, when it exists, the decomposition is unique (up to isomorphism and ordering of terms).
First, as you noted, $\{[1, +\infty), [2, 3], [2, 3]\}$ is not a correct interval decomposition for the persistence module in your question just by counting dimensions.
Second, using both $3$ and $+\infty$ as endpoints for the intervals is a little strange, since the two endpoints are the same in this case. In particular, $\mathbb{I}_{[1, \infty)}$ and $\mathbb{I}_{[1, 3]}$ are the same modules, so the other decompositions you proposed are all the same as the given one.
To avoid these kinds of confusion, let's write down the decomposition more explicitly. Since the composition of the two maps has rank one, there is one copy of $\mathbb{K} \to \mathbb{K} \to \mathbb{K}$ (all maps are identities) in the decomposition. The complement of this submodule is supported entirely on "$2$", so the whole decomposition is given by $$(\mathbb{K} \to \mathbb{K} \to \mathbb{K}) \oplus (0 \to \mathbb{K} \to 0) \oplus (0 \to \mathbb{K} \to 0).$$
The first module you might call $\mathbb{I}_{[1, 3]}$ (or $\mathbb{I}_{[1, \infty)})$. The second and third modules might be denoted $\mathbb{I}_{\{2\}}$ or perhaps $\mathbb{I}_{[2, 3)}$ (note the open right bracket). But at this point this is merely notation.
Finally, some remarks about uniqueness. The argument given above actually shows that this must be the interval decomposition of the module in question, so it is unique. More generally, intuitively, if you had two decompositions, you could intersect them to get a third decomposition that is finer than each. However, since interval modules are indecomposable, this finer decomposition must be the same as the original decompositions, hence the two original decompositions are equivalent. (This is essentially how the proof goes, but with more careful hypotheses.)