# Volume form of a submanifold with codimension $>1$

I have some problems with the definition of volume form of a submanifold with arbitrary condimension into some target Riemannian manifold.

Consider $$f:\Sigma^k\to M^n$$ an oriented immersion of a smooth manifold $$\Sigma$$ into a Riemannian manifold $$(M^n,g)$$, with $$2\leq k. The first question is how can I define the volume form of $$\Sigma$$. Can I copy the ideia of hypersurfaces, for example $$dvol_{\Sigma}(X_1,\cdots,X_k)=dvol_M(X_1,\cdots,X_k,\eta_1,\cdots,\eta_{m-k}),$$ where $$X_1,\cdots,X_k$$ are tangent vectors to $$\Sigma$$ and $$\eta_1,\cdots,\eta_{m-k}$$ normal vectors to $$\Sigma$$?

The second one is if we suppose that $$g$$ is conformally equivalent to $$\bar{g}$$, that means, $$g = \lambda \bar{g}$$, can I compute the volume form of $$\Sigma$$ in terms of $$\bar{g}$$? I think we will have some power of $$\lambda$$, but I don't know how relate this with the dimension of $$\Sigma$$.

I appreciate any help.

• You'd better specify that the normal vectors form an (oriented) orthonormal basis for the normal space. Commented Jul 30, 2020 at 18:35

(1) yes, as long as $$M$$ is orientable and your ordered (local) unit normal vectors are compatible. Note that the normal bundle is not necessarily trivializable, unlike the hypersurface case.
(2) If $$\bar{g}$$ is changed to $$g=\lambda\bar{g}$$, then the length of a tangent vector is scaled by $$\sqrt{\lambda}$$ and so the volume form on $$M$$: $$dvol_{g}=\lambda^{\dim M/2}dvol_{\bar{g}}$$ and on $$\Sigma$$: $$dvol_{\Sigma,g}=\lambda^{\dim\Sigma/2}dvol_{\Sigma,\bar{g}}$$.