Are all smooth functions composites of 0-, 1-, and 2-ary functions?

Question

I will formalize my question in terms of algebraic theories.

Background:

Recall that an algebraic theory (in the sense of Lawvere) is a category $\mathcal{C}$ which is closed under taking finite products, and whose set of objects can be identified with the set $\mathrm{Ob}(\mathcal{C})\cong\{T^0,T^1,\ldots\}$, where $T^i=T^1\times T^1\times\cdots\times T^1$ is the $i$-fold product of $T^1$. We denote $T^1$ by $T$. The interesting aspect of an algebraic theory $\mathcal{C}$ are the morphisms $\mathcal{C}(T^i,T^1)$ and their compositions.

For any algebraic theory $\mathcal{C}$, a $\mathcal{C}$-model is defined to be a functor $\mathcal{C}\to\mathbf {Set}$ that preserves all finite products. Denote the category of $\mathcal{C}$-models (and natural transformations between them) by $\mathbf{Mod}(\mathcal{C})$.

Let $n\in\mathbb{N}$ be a natural number, and let $\mathcal{C}$ be an algebraic theory. Its $n$-truncation, denoted $\mathcal{C}_{\leq n}\subseteq\mathcal{C}$, is the smallest algebraic sub-theory that has the same $k$-ary functions $$\mathcal{C}_n(T^k,T^1):=\mathcal{C}(T^k,T^1),$$ for all $k\leq n$. full subcategory spanned by the objects $\{T^0,\ldots,T^n\}$. We say that $\mathcal{C}$ is $n$-truncated if it is equivalent (as a category) to its $n$-truncation.

Setup:

Let $R$ denote the algebraic theory whose morphisms $T^n\to T$ are the smooth functions ${\mathbb R}^n\to{\mathbb R}$. This of course defines the set of morphisms $T^n\to T^m$ for any $m$. Endow $R$ with the usual formula for composing smooth functions.

To me it would be quite surprising if $R$ were 2-truncated. But I've never heard of a 3-ary function $\mathbb{R}^3\to\mathbb{R}$ that wasn't constructed from a combination of 0-, 1-, and 2-ary functions.

Question: Can you prove that $R$ is not 2-truncated? More interestingly, can you name, i.e., construct a 3-ary function that isn't also constructible by a combination of 0-, 1-, and 2-ary functions?

(Note: thanks to Todd Trimble for suggestions on how to clean up this question.)

Answer 1

For any $n$, there is an $n$-ary smooth function that is not a composition of smooth functions of lower arity; according to this answer to a very similar question, this is due to Vitushkin (at least for $n=3$). Here is a simple but rather inexplicit proof (adapted from this paper of Akashi and Kodama). Suppose that a smooth n-ary function $f$ can be expressed as a composition of smooth functions $g_i$ of lower arity (WLOG, all of arity $n-1$). Then for each $k$, the $k$-jet of $f$ at any point is determined by the $k$-jets of the $g_i$ at corresponding points. But dimension of the space of $n$-ary $k$-jets grows like $k^n$, while the dimension of the space of $(n-1)$-ary $k$-jets grows like $k^{n-1}$. It follows that for any particular way to compose lower arity functions to get an $n$-ary function, most $k$-jets of $n$-ary functions cannot be so obtained for $k$ sufficiently large. Since there are only countably many such ways to compose, we can use bump functions to construct a single $f$ that has jets at different points that fail for all of them. Actually, by a diagonal argument we can find a single $n$-ary $\infty$-jet (i.e., power series) that is not a composition of jets of lower arity, and every $\infty$-jet can be realized by a smooth function.