Volume form of a submanifold with codimension $>1$

Question

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 $2\leq k<n-1$. 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.

Answer 1

(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 $\Sigma$: $dvol_{\Sigma,g}=\lambda^{\dim\Sigma/2}dvol_{\Sigma,\bar{g}}$.