# Is $df$ exact for a map between manifolds?

Let $M,N$ be $d$-dimensional Riemannian manifolds, $f:M \to N$ a smooth map. Then $df \in \Omega^1\big({M,f^*TN}\big)=\Gamma(T^*M \otimes f^*TN)$.

Let $\nabla$ be the pullback connection on $f^*TN$ induced by the Levi-Civita connection on $TN$.

Does there exist an element $\sigma \in \Gamma(f^*TN)$ such that $\nabla \sigma=df$?

Denote by $d_{\nabla}:\Omega^k\big({M,f^*TN}\big) \to \Omega^{k+1}\big(M,f^*TN\big)$ the associated exterior derivative.

Then, if such a $\sigma$ exist, then $-R^{f^*TN} \wedge \sigma =d_{\nabla} d_{\nabla} \sigma=d_{\nabla} \nabla \sigma=d_{\nabla} df=0$, where the last equality comes from the symmetry of the connection on $TN$.

Thus,

$R^{f^*TN} \wedge \sigma =0$ is a necessary condition for such a $\sigma$. ($R^{f^*TN}$ is the curvature tensor of $\nabla$).

• If $d = 2$, $N$ has non-zero curvature and $f$ is a diffeomorphism, then for linearly independent $X,Y$ the endomorphism $R^{f^{*}(TN)}(X,Y)$ is invertible which forces $\sigma = 0$ but then $\nabla \sigma \neq df$. I'm not sure about the general case but I would guess that this fails generically. – levap Jun 12 '17 at 12:26
• Thanks. Can you elaborate on why $R^{f^*(TN)}(X,Y)$ is invertible? I see why it's non-zero (by your assumption). I guess the invertibility is a consequence of the $2$-dimensionality? (Something along "there is only one number determining the curvature tensor"? I wonder whether you have a slick argument for that.) – Asaf Shachar Jun 18 '17 at 16:43
• Yep, the endomorphism is skew-adjoint on a two dimensional space so it is invertible iff it is non-zero. – levap Jun 20 '17 at 17:56