Consider the following partial order. The objects are unordered tuples $\{V_1,\ldots,V_m\}$, where each $V_i \subseteq \mathbf{R}^n$ is a nontrivial linear subspace and $V_1 \oplus \cdots \oplus V_m = \mathbf{R}^n$. We say that $\{V_1,\ldots,V_m\} \geq \{W_1,\ldots,W_i\}$ if for each $j=1,\ldots,m$ there exists $1 \leq k \leq i$ such that $V_j \subseteq W_k$. (I.e. if $\{V_1,\ldots,V_m\}$ is a finer decomposition than $\{W_1,\ldots,W_i\}$.) This partial order has a minimal element, $\{\mathbf{R}^n\}$, which is the trivial decomposition. If we remove it, what is the homotopy type of this partial order?

Some things that are known about similar problems.

  1. If we're working over $\mathbb{F}_1$, i.e. with subsets of $\{1,\ldots,n\}$, then there is both a minimal and a maximal object. If we remove both of these then the homotopy type of the partial order is a wedge of $S^{n-3}$'s.

  2. If instead we work with ordered decompositions, then this is a barycentric subdivision of Ruth Charney's split building, so it is a wedge of $S^{n-2}$'s.

Edit: clarity and notation.

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    $\begingroup$ After work of Richard Stanley in the paper "Exponential Posets", the analogous problem with R replaced by a finite field was settled by Volkmar Welker in "Direct Sum Decompositions of Matroids and Exponential Structures. Welker's work uses shellability. In work of Anders Bj\"orner on shellability, a definition for infinite posets is given. You might be able to put all of this together to prove that the order complex is a wedge of (equidimensional) spheres. $\endgroup$ Jan 26, 2022 at 20:46
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    $\begingroup$ Are the $V_i$'s subvector spaces and the direct sum what it looks to be? Your notation makes this a bit unclear to me. If so, I'd look at the work of Arone and Lesh. $\endgroup$ Jan 26, 2022 at 23:38
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    $\begingroup$ Does $\mathbf R$ denote the real numbers, or a more general ring or field? If $\mathbf R$ denotes the real numbers, then the poset of direct-sum decompositions has a natural topology. Do you take the topology into account when you define the homotopy type? Notice that if $\mathbf R=$ the real numbers, and treat your poset as a discrete poset, then, assuming the geometric realization is spherical, it will be a wedge of uncountably many spheres. The topological posets, and their complex analogues, arise in the context of orthogonal calculus and also the rank filtration of K-theory. (continued) $\endgroup$ Jan 27, 2022 at 3:30
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    $\begingroup$ The topological geometric realization of these posets are (unsuprisingly) not wedges of spheres, but they are very interesting spaces with an action of $O(n)$, or, if you replace $\mathbf R$ with the complex numbers, an action of $U(n)$. The results are cleanest to state in the complex case: the topological poset of decompositions of $\mathbb C^n$ is rationally trivial for $n>1$, is contractible unless $n$ is a power of a prime, and if $n=p^k$ it is a variant of a finite complex with $A_k$-free mod $p$ cohomology that was constructed by Mitchell. $\endgroup$ Jan 27, 2022 at 3:36
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    $\begingroup$ To clarify something from the comments: "homotopy type of a poset" is usually understood to mean "homotopy type of the simplicial complex whose faces are chains in the poset" (i.e., the homotopy type of the "order complex" of the poset). $\endgroup$ Jan 27, 2022 at 22:17

1 Answer 1


Let me write $V$ for a finite-dimensional vector space over some field (the field will not play a role), and $\mathsf{P}(V)$ for the poset described in the question, which I consider as a category. Let me rephrase the order relation: $\{V_1, \ldots, V_n\} \geq \{W_1, \ldots, W_m\}$ if and only if each $W_j$ is a direct sum of $V_i$'s. The nerve of this poset is $(\mathrm{dim}(V)-2)$-dimensional, so to see that it is a wedge of spheres one must show that it is $(\mathrm{dim}(V)-3)$-connected. I think this is true.

Let $\mathsf{S}(V)$ denote the following category. It has objects given by pairs $([n], f : [n] \to Sub(V))$ consisting of a standard set $[n] := \{1,2,\ldots, n\}$ and a function from $[n]$ to the set of proper vector subspaces of $V$, such that $\bigoplus_{i \in [n]} f(i) = V$. A morphism in $\mathsf{S}(V)$ from $([n], f)$ to $([n'], f')$ is given by a surjection $e : [n] \to [n']$ such that $f'(i) = \bigoplus_{j \in e^{-1}(i)} f(j)$.

There is a functor $F : \mathsf{S}(V) \to \mathsf{P}(V)$ given by sending $([n], f)$ to the unordered collection $\{f(i)\}_{i \in [n]}$.

Now let me do something a bit odd: choose a total order $\prec$ on the set of vector subspaces of $V$. Then we can consider objects of $\mathsf{P}(V)$ as given by lists of subspaces $(V_1, \ldots, V_n)$ with $V_1 \prec V_2 \prec \cdots \prec V_n$, and such that $\bigoplus_{i=1}^n V_i = V$. Attempt to define a functor $G : \mathsf{P}(V) \to \mathsf{S}(V)$ by $$G(V_1, \ldots, V_n) := ([n], i \mapsto V_i)$$ on objects. If $(V_1, \ldots, V_n) \geq (W_1, \ldots, W_m)$ in $\mathsf{P}(V)$ then each $W_j$ is a direct sum of $V_i$'s. Define a function $e : [n] \to [m]$ by $e(i) = j$ if $V_i \subset W_j$; this gives a morphism in $\mathsf{S}(V)$ from $([n], i \mapsto V_i)$ to $([n], j \mapsto W_j)$. To check that this is indeed a functor suppose that $(W_1, \ldots, W_m) \geq (U_1, \ldots, U_\ell)$, with $e'(j) = k$ when $W_j \subset U_k$. Then we indeed have $e' e(i) = k$ when $V_i \subset W_{e(i)} \subset U_k$, so $G$ is indeeed a functor.

The conclusion of this discussion is that $\mathsf{P}(V)$ is a retract of $\mathsf{S}(V)$, so it suffices to show that $\mathsf{S}(V)$ is $(\mathrm{dim}(V)-3)$-connected.

The full name of $\mathsf{S}(V)$ is $\mathsf{S}^{E_\infty}(V)$, the $E_\infty$-splitting category defined by Galatius, Kupers, and I in Definition 17.17 of Cellular $E_k$-algebras (applied to the symmetric monoidal groupoid of finite-dimensional vector spaces; the way I have described it here is not identical to that definition, but is an equivalent category). It has a cousin $\mathsf{S}^{E_1}(V)$ which turns out to be precisely (the category of simplices of) Charney's split building. In particular $\mathsf{S}^{E_1}(V)$ is $(\mathrm{dim}(V)-3)$-connected by Charney's theorem.

It would take too long to explain all the details here, but the theory developed in that paper shows that the collection of all $\Sigma^2 \mathsf{S}^{E_\infty}(V)$'s can be obtained from the collection of all $\Sigma^2 \mathsf{S}^{E_1}(V)$'s by an iterated bar construction, and in particular given Charney's connectivity result (for all $V$) it follows that the $\Sigma^2 \mathsf{S}^{E_\infty}(V)$ are also $(\mathrm{dim}(V)-1)$-connected, so the $\mathsf{S}^{E_\infty}(V)$ are homologically $(\mathrm{dim}(V)-3)$-connected. (It should not be hard to show it is simply-connected, by hand.) I'm happy to discuss the details by e-mail, if you like.

(Also, now that I believe it is true I expect there must be a more elementary way to deduce it from Charney's theorem.)


  1. As @inna says in the comments, the functor $F : \mathsf{S}(V) \to \mathsf{P}(V)$ is actually an equivalence, because there is a natural isomorphism $\mathrm{Id} \Rightarrow F \circ G$ given by applying the unique permutation necessary to put things in the order given by $\prec$.

  2. Let me try to give some references for what I said in the final paragraph. All references are to Cellular $E_k$-algebras:

Writing $\mathsf{G}$ for the symmetric monoidal category of finite-dimensional vector spaces and linear isomorphisms, we work in the category $\mathsf{sSet}_*^\mathsf{G}$ of functors from $\mathsf{G}$ to pointed simplicial sets. This is again symmetric monoidal by Day convolution, and the functor $$\mathbb{t}(V) = \begin{cases} S^0 & V \neq 0\\ * & V=0 \end{cases}$$ is a nonnital commutative monoid, and hence also a nonunital $E_\infty$-algebra. (In the paper this object is called $\underline{*}_{>0}$.)

There is another character involved: the derived $E_k$-indecomposables $Q^{E_k}_\mathbb{L}(\mathbb{t})$ for $1 \leq k \leq \infty$, which are again objects of $\mathsf{sSet}_*^\mathsf{G}$. For $V$ an object of $\mathsf{G}$ we write $H_{V,d}^{E_k}(\mathbb{t}) := H_d(Q^{E_k}_\mathbb{L}(\mathbb{t})(V))$.

The ingredients I have in mind are now:

a. Combining Proposition 17.4 and Lemma 17.10 shows that $$\Sigma Q^{E_1}_\mathbb{L}(\mathbb{t})(V) \simeq \Sigma^2 \mathsf{S}^{E_1}(V)$$ (in fact by Section 17.5 this can be desuspended once) and hence Charney's theorem shows that $H_{V,d}^{E_1}(\mathbb{t})=0$ for $d < \dim(V)-1$.

b. Theorem 14.4 (this is the application of the bar construction result, but is packaged so one doesn't explicitly have to think about that) applied with $\rho(V) := \mathrm{dim}(V)$ shows that $H_{V,d}^{E_\infty}(\mathbb{t})=0$ for $d < \dim(V)-1$ too.

c. Corollary 17.23 shows that $Q^{E_\infty}_\mathbb{L}(\mathbb{t})(V) \simeq \Sigma \mathsf{S}^{E_\infty}(V)$ so the above translates to $\tilde{H}_*(\mathsf{S}^{E_\infty}(V))=0$ for $* \leq \mathrm{dim}(V)-3$.

  • $\begingroup$ This looks awesome. I'll look at it a bit more closely, but I think this will work. Could you maybe give me a reference to the bar construction result? $\endgroup$
    – Inna
    Jan 29, 2022 at 23:53
  • $\begingroup$ Also: I think that it should be deducible from Charney's result (that's why I mentioned in in #2 above) but I haven't been able to figure out a way to do this. $\endgroup$
    – Inna
    Jan 29, 2022 at 23:54
  • $\begingroup$ I think that your definition of $S(V)$ is equivalent to $P(V)$, just a different way of stating the objects and definitions. Am I missing something? $\endgroup$
    – Inna
    Jan 30, 2022 at 1:10
  • $\begingroup$ I think maybe it is indeed "easy" to see it's a retract of Charney's complex. Take the partial order $\le$ on proper non-trivial subspaces of V, and extend to a total order $\preceq$. (So $\preceq$ is not an arbitrary total order, it respects inclusion of subspaces.) Now define an embedding $\iota$ of the (unordered) complex into Charney's (ordered) complex by sending $\{V_1,\dots,V_m\}$ to $(V_1,\dots,V_m)$ where $V_1\prec \cdots \prec V_m$. I think this induces a simplicial map since $\preceq$ extends $\le$. And the "forget the order" map $\rho$ has $\rho\circ\iota=id$, so it's a retract. $\endgroup$ Jan 30, 2022 at 1:11
  • $\begingroup$ @MattZaremsky: I think this will not be a map of posets. In Charney's poset the order relation is induced by taking direct sum of contiguous sequences of subspaces. So if $V_1 \prec V_2 \prec V_3$ then there is a map $\{V_1, V_2, V_3\} \geq \{V_1 \oplus V_3, V_2\}$ in the unordered poset, but in the ordered poset the only things under $(V_1, V_2, V_3)$ are $(V_1 \oplus V_2, V_3)$ and $(V_1, V_2 \oplus V_3)$. $\endgroup$ Jan 30, 2022 at 7:38

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