# Is $C^{\infty}(\mathbb{R}^{m+n})$ a flat module over $C^{\infty}(\mathbb{R}^{m})$?

For $$m>0$$ we consider the ring $$C^{\infty}(\mathbb{R}^{m})$$ of smooth functions on $$\mathbb{R}^{m}$$. For $$n>0$$ we consider the projection $$\mathbb{R}^{m+n}\to \mathbb{R}^{m}$$ hence $$C^{\infty}(\mathbb{R}^{m+n})$$ is naturally a $$C^{\infty}(\mathbb{R}^{m})$$ module.

My question is: is it true that $$C^{\infty}(\mathbb{R}^{m+n})$$ is a flat module over $$C^{\infty}(\mathbb{R}^{m})$$? I'm not sure if we should take the topology on algebra into consideration.

• Is it perhaps even a free module? I have no intuition one way or another. – Gro-Tsen Dec 15 '19 at 16:50
• My suspicion is that this should work if one uses the topological variants of flatness etc set out in the work of Helemskii and his school -- there might be some relevant work of Ogneva -- but I am away from my reference sources right now – Yemon Choi Dec 15 '19 at 21:10
• "Example-Lemma 4.30" of springer.com/gp/book/9780387955438 might be related. – Qfwfq Dec 15 '19 at 21:15
• Actually, I see that in an answer to one of your previous questions mathoverflow.net/a/304075 Igor Khavkhine already mentioned Ogneva's work, with a link to the relevant paper(s) and some technical details – Yemon Choi Dec 15 '19 at 21:26

This is just a long comment. Let $$X,Y$$ be smooth manifolds (here in particular, resp. $$\mathbb{R}^n$$ and $$\mathbb{R}^m$$), and let $$M$$ be a $$C^\infty(X)$$-module. Then any pair $$(m,\phi)\in M\times C^\infty(X\times Y)$$ defines a function $$Y\to M$$, namely $$Y\ni c\mapsto \phi(x,c)m\in M$$ (which is well defined as for all $$c\in Y$$, $$\phi(x,c)\in C^\infty(X)$$). Since this is correspondence is bilinear $$M\times C^\infty(X,Y) \to M^Y$$, it produces a morphism of $$C^\infty(X)$$-modules $$M\otimes C^\infty(X,Y) \to M^Y.$$ Is this morphism injective? That is, can one see $$M\otimes C^\infty(X,Y)$$ as a space of $$M$$-valued functions on $$Y$$? If the answer is yes, then any morphism $$f:M\to M'$$ produces a $$f\otimes 1:M\otimes C^\infty(X,Y)\to M'\otimes C^\infty(X,Y)$$ which extends to the morphism $$M^Y\ni u\mapsto f\circ u\in M'^Y$$, which is injective whenever $$f$$ is injective. Therefore $$C^\infty(X,Y)$$ would be a flat $$C^\infty(X)$$-module.

PS: The answer below has gaps, and it is likely incorrect.

Yes, $$C^{\infty}(\mathbb{R}^{m+n})$$ is a flat $$C^{\infty}(\mathbb{R}^{m})$$ module. Or, following @Pietro 's comment, with more generality, if $$X,Y$$ are smooth manifolds, then $$C^\infty(X\times Y)$$ is a $$C^\infty(X)$$ flat module. To see this, it is enough to consider $$C^\infty(X)$$ linear, injective maps $$\varphi: K \rightarrow L$$, where $$K = _{C^\infty(X)}$$ is a finitely generated $$C^\infty(X)$$ ideal, and $$L=C^\infty(X)$$ is the ring of scalars.

Notice that in that case, considering that $$C^\infty(X)$$ is a sub-ring of $$C^\infty(X\times Y)$$, the tensor operation $$.\otimes_{C^\infty(X)} C^\infty(X\times Y)$$ applied to an ideal such as $$K$$ is just the extension of scalars, i.e., in some natural way, $$K\otimes_{C^\infty(X)} C^\infty(X\times Y) = _{C^\infty(X\times Y)},$$ and $$L\otimes_{C^\infty(X)} C^\infty(X\times Y) = C^\infty(X\times Y).$$ Moreover, with these representations, $$\bar{\varphi} = \varphi \otimes Id$$, the extension of $$\varphi$$ from $$K$$ to $$K\otimes C^\infty(X\times Y)$$, satisfies: $$\bar{\varphi}( \sum_{i=1}^d r_i\ m_i) = \sum_{i=1}^d \varphi(r_i)\ m_i$$ for any $$m_1,\dots, m_d \in C^\infty(X\times Y)$$.

We want to verify that, under the assumption the $$\varphi$$ is injective, $$\bar{\varphi}$$ is injective. Now, if for some linear combination $$\sum_i r_i \ m_i$$ we have $$\bar{\varphi}(\sum_i r_im_i) = 0$$, then $$\sum_i \varphi(r_i)\ m_i = 0$$ in $$C^\infty(X\times Y)$$, and so, for all $$y \in Y$$, $$\sum_i \varphi(r_i)\ m_i(.,y) = 0$$ in $$C^\infty(X)$$, but $$\varphi$$ being $$C^\infty(X)$$ linear, we get: $$\varphi( \sum_i r_i\ m_i(.,y) ) = 0.$$ By injectivity of $$\varphi$$, $$\sum_i r_i\ m_i(.,y) = 0$$

(With @Pietro notation, this is the condition of an element of $$M^Y$$ having all 0 coordinates applied to the case $$M =$$ a finitely generated ideal of $$C^\infty(X)$$)

Since this holds for all $$y$$, then $$\sum_i r_i\ m_i = 0$$, which shows injectivity of $$\varphi$$.

• I am outside my comfort zone when it comes to Frechet algebras and modules rather than Banach ones, but in extension-of-scalars arguments don't we need to ise a suitable completed tensor product? Even at the vector space level, my gut feeling is that $C^\infty({\bf R})\otimes C^\infty({\bf R})$ is all of $C^\infty({\bf R}^2)$, as this seems similar to claiming that all continuous linear maps $C^\infty({\bf R})\to C^\infty ({\bf R})'$ have finite rank... – Yemon Choi Dec 15 '19 at 21:09
• @Yemon Choi: According to Nestruev (see my previous comment in the OP), $\mathcal{C}^\infty (M\times N)$ is isomorphic to the "smooth envelope" (3.36) of the tensor product over $\mathbb{R}$ of the two algebras. There, "$\mathcal{C}^\infty$-closed" is defined by means of a property (3.32) that reminds me of the $\mathcal{C}^\infty$-schemes by Joyce. – Qfwfq Dec 15 '19 at 21:49
• Maybe I'm missing some obvious fact, but the last line is not quite clear to me: If, for all $y\in Y$, one has $\sum_i r_i m_i(.,y)=0$ as an element of $K$, that is $\sum_i r_i m_i$ induces the zero function, that is it is $0$ as an element of $K^Y$, why can we conclude that $\sum_i r_i m_i=0$ as an element of $K\otimes C^\infty(X\times Y)$? – Pietro Majer Dec 15 '19 at 23:50
• Translating into algebraic language: Set $R := C(X), S := C(X\times Y), \varphi: K \hookrightarrow R$ for $K \triangleleft R$. Then $R \le S$ and what you are saying is that there is a commutative diagramm $$K \otimes_R S \xrightarrow{\varphi \otimes id} R\otimes_R S$$ $$\cong\, \downarrow \qquad\qquad \downarrow\ \,\cong$$ $$KS \quad \xrightarrow[\bar{\varphi}] \quad S$$ Why is the left vertical arrow an isomorphism? – tj_ Dec 16 '19 at 5:27
• I don't see how your argument uses in any way properties of the ring of smooth functions. Is the corresponding question known for rings of continuous functions on (locally) compact Hausdorff spaces $X$ and $Y$? This is perhaps in the comfort zone of Yemon Choi. – Jochen Wengenroth Dec 16 '19 at 9:46

Let $$X,Y$$ be smooth manifolds; let $$\Im = _{C^\infty(X)}$$ be a finitely generated ideal of $$C^\infty(X)$$. Then $$\Im \otimes_{C^\infty(X)} C^\infty(X \times Y) = \{ \sum_{i=1}^d r_i \otimes g_i \ | \ g_1,\dots,g_d \in C^\infty(X \times Y)\}$$. I claimed elsewhere that the condition $$\sum_{i=1}^d r_i \otimes g_i = 0$$ in $$\Im \otimes C^\infty(X \times Y)$$ is equivalent to $$\sum_{i=1}^d r_i g_i = 0$$ in $$C^\infty(X \times Y)$$. That is not the case. The correct statement is:

The following conditions are equivalent:

i) $$\sum_{i=1}^d r_i g_i = 0$$ in $$C^\infty(X \times Y)$$
ii) $$k(\sum_{i=1}^d r_i \otimes g_i) = 0$$ for all $$k \in \Im$$

Proof $$i) \Rightarrow ii)$$ If $$\sum_{i=1}^d r_i g_i = 0$$ and $$k \in \Im$$ then $$0 = k \otimes (\sum_{i=1}^d r_i g_i ) = \sum_{i=1}^d k \otimes (r_i g_i) =$$
$$\sum_{i=1}^d r_i \ (k \otimes g_i) = \sum_{i=1}^d (r_i k) \otimes g_i = k (\sum_{i=1}^d r_i \otimes g_i)$$

$$ii) \Rightarrow i)$$ Take $$k = r_j$$, and assume $$0 = r_j( \sum_{i=1}^d r_i \otimes \ g_i) =\sum_{i=1}^d r_i \otimes ( r_j \ g_i)$$. Fix $$a \in Y$$, and consider the bilinear map $$e : \Im \times C^\infty(X \times Y) \to C^\infty(X)$$ defined as $$e(k,g) = k(x)g(x,a)$$. If $$\bar{e}$$ is the corresponding map on $$\Im \otimes C^\infty(X \times Y)$$, then $$0 = \bar{e}( \sum_{i=1}^d r_i \otimes ( r_j \ g_i) ) = \sum_{i=1}^d \bar{e}(r_i \otimes ( r_j \ g_i)) =$$
$$\sum_{i=1}^d r_i \ r_j \ g_i(.,a) = r_j ( \sum_{i=1}^d r_i \ g_i(.,a) )$$

Since this holds for all $$j$$, at any point $$x \in X$$ where some $$r_j$$ does not vanish, $$\sum_{i=1}^d r_i(x) \ g_i(x,a) =0$$, and the same holds trivially at any point $$x$$ where all $$r_j$$'s vanish.

• As far as I see this does not answer question but rather shows that and why your previous answer is not conclusive. I would suggest that you edit your first answer and delete this one. – Jochen Wengenroth Dec 18 '19 at 8:27