In linear algebra and differential geometry, there are various structures which we calculate with in a basis or local coordinates, but which we would like to have a meaning which is basis independent or coordinate independent, or at least, changes in some covariant way under changes of basis or coordinates. One way to ensure that our structures adhere to this principle is to give their definitions without reference to a basis. Often we employ universal properties, functors, and natural transformations to encode these natural, coordinate/basis free structures. But the Riemannian volume form does not appear to admit such a description, nor does its pointwise analogue in linear algebra.

Let me list several examples.

  • In linear algebra, an inner product on $V$ is an element of $\operatorname{Sym}^2{V^*}$. The symmetric power is a space which may be defined by a universal property, and constructed via a quotient of a tensor product. No choice of basis necessary. Alternatively an inner product can be given by an $n\times n$ symmetric matrix. The correspondence between the two alternatives is given by $g_{ij}=g(e_i,e_j)$. Calculations are easy with this formulation, but one should check (or require) that the matrix transforms appropriately under changes of basis.

  • In linear algebra, a volume form is an element of $\Lambda^n(V^*)$. Alternatively one may define a volume form operator as the determinant of the matrix of the components of $n$ vectors, relative to some basis.

  • In linear algebra, an orientation is an element of $\Lambda^n(V^*)/\mathbb{R}^>$.

  • In linear algebra, a symplectic form is an element of $\Lambda^2(V^*)$. Alternatively may be given as some $\omega_{ij}\,dx^i\wedge dx^j$.

  • In linear algebra, given a symplectic form, a canonical volume form may be chosen as $\operatorname{vol}=\omega^n$. This operation can be described as a natural transformation $\Lambda^2\to\Lambda^n$. That is, to each vector space $V$, we have a map $\Lambda^2(V)\to\Lambda^n(V)$ taking $\omega\mapsto \omega^n$ and this map commutes with linear maps between spaces.

  • In differential geometry, all the above linear algebra concepts may be specified pointwise. Any smooth functor of vector spaces may be applied to the tangent bundle to give a smooth vector bundle. Thus a Riemannian metric is a section of the bundle $\operatorname{Sym}^2{T^*M}$, etc. A symplectic form is a section of the bundle $\Lambda^2(M)$, and the wedge product extends to an operation on sections, and gives a symplectic manifold a volume form. This is a global operation; this definition of a Riemannian metric gives a smoothly varying inner product on every tangent space of the manifold, even if the manifold is not covered by a single coordinate patch

  • In differential geometry, sometimes vectors are defined as $n$-tuples which transform as $v^i\to \tilde{v}^j\frac{\partial x^i}{\partial \tilde{x}^j}$ under a change of coordinates $x \to \tilde{x}$. But a more invariant definition is to say a vector is a derivation of the algebra of smooth functions. Cotangent vectors can be defined with a slightly different transformation rule, or else invariantly as the dual space to the tangent vectors. Similar remarks hold for higher rank tensors.

  • In differential geometry, one defines a connection on a bundle. The local coordinates definition makes it appear to be a tensor, but it does not behave the transformation rules set forth above. It's only clear why when one sees the invariant definition.

  • In differential geometry, there is a derivation on the exterior algebra called the exterior derivative. It may be defined as $d\sigma = \partial_j\sigma_I\,dx^j\wedge dx^I$ in local coordinates, or better via an invariant formula $d\sigma(v_1,\dotsc,v_n) = \sum_i(-1)^iv_i(\sigma(v_1,\dotsc,\hat{v_i},\dotsc,v_n)) + \sum_{i+j}(-1)^{i+j}\sigma([v_i,v_j],v_1,\dotsc,\hat{v_i},\dotsc,\hat{v_j},\dotsc,v_n)$

  • Finally, the volume form on an oriented inner product space (or volume density on an inner product space) in linear algebra, and its counterpart the Riemannian volume form on an oriented Riemannian manifold (or volume density form on a Riemannian manifold) in differential geometry. Unlike the above examples which all admit global basis-free/coordinate-free definitions, we can define it only in a single coordinate patch or basis at a time, and glue together to obtain a globally defined structure. There are two definitions seen in the literature:

    1. choose an (oriented) coordinate neighborhood of a point, so we have a basis for each tangent space. Write the metric tensor in terms of that basis. Pretend that the bilinear form is actually a linear transformation (this can always be done because once a basis is chosen, we have an isomorphism to $\mathbb{R}^n$ which is isomorphic to its dual (via a different isomorphism than that provided by the inner product)). Then take the determinant of resulting mutated matrix, take the square root, multiply by the wedge of the basis one-forms (the positive root may be chosen in the oriented case; in the unoriented case, take the absolute value to obtain a density).
    2. Choose an oriented orthonormal coframe in a neighborhood. Wedge it together. (Finally take the absolute value in the unoriented case).

Does anyone else think that one of these definitions sticks out like a sore thumb? Does it bother anyone else that in linear algebra, the volume form on an oriented inner product space doesn't exist as natural transformation $\operatorname{Sym}^2 \to \Lambda^n$? Do the instructions to "take the determinant of a bilinear form" scream out to anyone else that we're doing it wrong? Does it bother anyone else that in Riemannian geometry, in stark contrast to the superficially similar symplectic case, the volume form cannot be defined using invariant terminology for the whole manifold, but rather requires one to break the manifold into patches, and choose a basis for each? Is there any other structure in linear algebra or differential geometry which suffers from this defect?

Answer: I've accepted Willie Wong's answer below, but let me also sum it up, since it's spread across several different places. There is a canonical construction of the Riemannian volume form on an oriented vector space, or pseudoform on a vector space. At the level of level of vector spaces, we may define an inner product on the dual space $V^*$ by $\tilde{g}(\sigma,\tau)=g(u,v)$ where $u,v$ are the dual vectors to $\sigma,\tau$ under the isomorphism between $V,V^*$ induced by $g$ (which is nondegenerate). Then extend $\tilde{g}$ to $\bigotimes^k V^*$ by defining $\hat{g}(a\otimes b\otimes c,\dotsb,x\otimes y\otimes z\dotsb)=\tilde{g}(a,x)\tilde{g}(b,y)\tilde{g}(c,z)\dotsb$. Then the space of alternating forms may be viewed as a subspace of $\bigotimes^k V^*$, and so inherits an inner product as well (note, however that while the alternating map may be defined canonically, there are varying normalization conventions which do not affect the kernel. I.e. $v\wedge w = k! Alt(v\otimes w)$ or $v\wedge w = Alt(v\otimes w)$). Then $\hat{g}(a\wedge b\dotsb,x\wedge y\dotsb)=\det[\tilde{g}(a,x)\dotsc]$ (with perhaps a normalization factor required here, depending on how Alt was defined).

Thus $g$ extends to an inner product on $\Lambda^n(V^*)$, which is a 1 dimensional space, so there are only two unit vectors, and if $V$ is oriented, there is a canonical choice of volume form. And in any event, there is a canonical pseudoform.

  • $\begingroup$ +1: Good observations. I'd like to know the answer to this too. For the last part: the metric tensor induces an isomorphism between the tangent and cotangent bundles, so with enough abuse of isomorphisms, we can make the coordinate-free definition of determinant work here... but there's no conceptual benefit over thinking of it as a matrix. $\endgroup$
    – Zhen Lin
    Oct 8, 2011 at 21:40
  • $\begingroup$ We can use isomorphisms of one non degenerate bilinear form to convert another bilinear form into a linear transformation, but if the two bilinear forms are the same, then the linear transformation is the identity, and so contains no information, and certainly cannot tell you the "determinant" of the bilinear form. $\endgroup$
    – ziggurism
    Oct 9, 2011 at 17:05
  • $\begingroup$ Obviously. But the vector space in question comes with a natural basis, namely the coordinate basis, and so is already equipped with an isomorphism to its dual space. $\endgroup$
    – Zhen Lin
    Oct 9, 2011 at 20:57
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    $\begingroup$ This is the non-degenerate bilinear form that provides the canonical ("musical") isomorphisms between the spaces of vectors and 1-forms, that you can use now to identify the spaces of bilinear forms and self-adjoint linear transformations. This identification allows you to define the determinant of a bilinear form as the determinant of the corresponding linear transformation. $\endgroup$ Oct 9, 2011 at 21:59
  • 1
    $\begingroup$ For the definition to be basis-free you don't need a "natural" basis, but independence of a choice of a basis. This is achieved by using "canonical" (i.e. free of choice) isomorphisms. What you call "natural" basis in $\mathbb{R^n}$ is just a convenient CHOICE of a basis, so called the standard basis in $\mathbb{R^n}$. $\endgroup$ Oct 9, 2011 at 22:09

3 Answers 3


A few points:

  • It is necessary to define "Riemannian volume forms" a patch at a time: you can have non-orientable Riemannian manifolds. (Symplectic manifolds are however necessarily orientable.) So you cannot just have a global construction mapping Riemannian metric to Riemannian volume form. (Consider the Möbius strip with the standard metric.)
  • It is however to possible to give a definition of the Riemannian volume form locally in a way that does not depend on choosing a coordinate basis. This also showcases why there cannot be a natural map from $\mathrm{Sym}^2\to \Lambda^n$ sending inner-products to volume forms. We start from the case of the vector space. Given a vector space $V$, we know that $V$ and $V^*$ are isomorphic as vector spaces, but not canonically so. However if we also take a positive definite symmetric bilinear form $g\in \mathrm{Sym}_+^2(V^*)$, we can pick out a unique compatible isomorphism $\flat: V\to V^*$ and its inverse $\sharp: V^*\to V$. A corollary is that $g$ extends to (by abuse of notation) an element of $\mathrm{Sym}_+^2(V)$. Then by taking wedges of $g$ you get that the metric $g$ (now defined on $V^*$) extends to uniquely to a metric1 on $\Lambda^k(V^*)$. Therefore, up to sign there is a unique (using that $\Lambda^n(V^*)$ is one-dimensional) volume form $\omega\in \Lambda^n(V^*)$ satisfying $g(\omega,\omega) = 1$. But be very careful that this definition is only up to sign.
  • The same construction extends directly to the Riemannian case. Given a differentiable manifold $M$. There is a natural map from sections of positive definite symmetric bilinear forms on the tangent space $\Gamma\mathrm{Sym}_+^2(T^*M) \to \Gamma\left(\Lambda^n(M)\setminus\{0\} / \pm\right)$ to the non-vanishing top forms defined up to sign. From which the usual topological arguments shows that if you fix an orientation (either directly in the case where $M$ is orientable or lifting to the orientable double cover if not) you get a map whose image now is a positively oriented volume form.

Let me just summarise by giving the punch line again:

For every inner product $g$ on a vector space $V$ there are two compatible volume forms in $\Lambda^n V$: they differ by sign. Therefore the natural mapping from inner products takes image in $\Lambda^n V / \pm$!

Therefore if you want to construct a map based on fibre-wise operations on $TM$ sending Riemannian metrics to volume forms, you run the very real risk that, due to the above ambiguity, what you construct is not even continuous anywhere. The "coordinate patch" definition has the advantage that it sweeps this problem under the rug by implicitly choosing one of the two admissible local (in the sense of open charts) orientation. You can do without the coordinate patch if you start, instead, with an orientable Riemannian manifold $(M,g,\omega)$ and use $\omega$ to continuously choose one of the two admissible pointwise forms.

1: this used to be linked to a post on MathOverflow, which has since been deleted. So for completeness: the space of $k$-tensors is the span of tensors of the form $v_1 \otimes \cdots \otimes v_k$, and you can extend $g$ to the space of $k$-tensors by setting $$ g(v_1\otimes\cdots v_k, w_1\otimes\cdots\otimes w_k) := g(v_1, w_1) g(v_2, w_2) \cdots g(v_k, w_k) $$ and extending using bilinearity. The space $\Lambda^k(V^*)$ embeds into $\otimes^k V^*$ in the usual way and hence inherits a inner product.

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    $\begingroup$ As a corollary: on a orientable Riemannian manifold, there isn't just one, but two Riemannian volume forms. However, on a Kahler manifold, one of the two Riemannian volume forms is not compatible with the symplectic one. $\endgroup$ Nov 30, 2011 at 9:38
  • $\begingroup$ Your excellent answer has reminded me of the issue with orientations. So I should never have expected a natural transformation from $\operatorname{Sym}^2 \to \Lambda^n$ where the codomain is vector spaces. I should either ask for a natural transformation between the corresponding functors defined on oriented vector spaces, or else a natural map $\operatorname{Sym}^2 \to |\Lambda^n|$, the densities functor. The ambiguity you mention will disappear with either modification, right? So I can still ask for a natural, perhaps global, construction? I will update the question with this change. $\endgroup$
    – ziggurism
    Nov 30, 2011 at 13:39
  • $\begingroup$ So when you write "Then by taking wedges of $g$ you get that the metric $g$ [..] extends uniquely to a metric on $\Lambda^k(V^∗)$", that's the construction where you take the factors of a pair of $k$-vectors, and fill up a matrix with $g$ acting on those factors, and then take the determinant, right? I'd like to look closely at that construction, it seems to be the essence of my question. How do we decompose an arbitrary $k$-vectors into its factor 1-vectors? Can that be done invariantly? $\endgroup$
    – ziggurism
    Nov 30, 2011 at 14:00
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    $\begingroup$ Decomposing an arbitrary $k$-vector into its factor 1-vectors is the same as identifying a basis of $\otimes^k V$ from a basis of $V$. You should read MTS's comment (and the link to another of his answers in that comment) on the MO thread I linked to about extending $g$ to $\Lambda^kV$. $\endgroup$ Nov 30, 2011 at 16:37
  • $\begingroup$ Yes, Willie, that is the complete explanation I was looking for. Thank you! And thanks to @MTS by proxy. Exercises remaining include extending the definition of $g$ to densities, and showing that this construction is indeed a natural transformation. $\endgroup$
    – ziggurism
    Nov 30, 2011 at 21:42

A coordinate-free definition of volume form is in fact well-known and frequently used, e.g. the cited Wikipedia article. I will try to reproduce it the nutshell to the best of my understanding.

Let $V$ be a (real, for certainty) vector space of finite dimension $\dim V = n$. The space of $n$-forms $\Lambda^n (V)$ has dimension 1. Thus $\Lambda^n (V)$ isomorphic to $\mathbb{R}$, however this isomorphism is not canonical: any choice of non-trivial $n$-form $\omega$ can be mapped to $1 \in \mathbb{R}$.

A volume form on a finite-dimensional vector space $V$ is a choice of a top-rank non-trivial exterior form (skew-symmetric $n$-linear functional) $\omega \in \Lambda^n (V)$. I think that this definition is quite coordinate-free.

Once such a form has been chosen, it can be used to divide the space of bases in $V$ into two classes that are called orientations. There are two of them, positive ($\omega > 0$) and negative ($\omega < 0$). Having a volume form chosen, one can speak about oriented volumes of parallelotopes, for instance.

If for any reason we have an inner product $g$ in $V$ we can make this choice canonical. One needs to consider orthonormal frames (with respect to $g$). The canonical volume form will take value 1 on positively oriented orthonormal frames.

The volume form of an inner-product space $(V, g)$ is that canonical choice of a volume form. It can be denoted by $Vol_{g}$ provided one also keeps in mind that there is a choice of orientation involved.

Along these lines one can obtain an understanding of the volume form as the Hodge dual of 1 in a pretty coordinate-free manner.

  • $\begingroup$ The definition of a volume form on an arbitrary vector space is nice: it's an element of $\Lambda^n(V^*)$. That space can be defined via a universal property, and constructed via a quotient of a cartesian product. Never any reference to a basis. $\endgroup$
    – ziggurism
    Nov 30, 2011 at 4:02
  • $\begingroup$ A symplectic vector space is defined by choosing an element $\omega$ from $\Lambda^2(V^*)$, without specifying a basis, or saying how it acts on bases. You can get a volume form from $\omega$ by just doing a linear operation. That operation, viz. the wedge product, is defined without reference to a basis either via a universal property or else as a projection to a quotient. $\endgroup$
    – ziggurism
    Nov 30, 2011 at 4:02
  • $\begingroup$ The inner product $g$ is an element of $\operatorname{Sym}^2(V^*)$. To get a volume form from it, I have to specify a basis compatible with $g$ and then specify how my volume form acts on it. This seems like a much uglier definition to me! I want there to be some fancy multilinear operation from $\operatorname{Sym}^2(V)\to \Lambda^n(V)$. I guess if such a thing existed, it would be in the textbooks? $\endgroup$
    – ziggurism
    Nov 30, 2011 at 4:03
  • $\begingroup$ I went to work some of these comments into the original question, but ended up re-writing the question from top to bottom. Hopefully my issue will be clearer? $\endgroup$
    – ziggurism
    Nov 30, 2011 at 5:29
  • $\begingroup$ I think: one can do coordinate free when you just do vector space, but OP (@ziggurism) is asking for manifold? $\endgroup$
    – BCLC
    Dec 31, 2020 at 12:58

The metric tensor is symmetric and so can be orthogonally diagonalized, which means that its determinant is just the product, with multiplicity, of its eigenvalues. If we rotate the entire chart such that the coordinate axes at the point of interest follow the the eigenspaces, this means that $|g|$ tells us the factor that relates the squared coordinate length of each side of an infinitesimal box with its squared intrinsic length, all multiplied together. Therefore, multiplying with the square root of the determinant will convert a coordinate volumne of an infinitesimal box to the intrinsic volume.

The intuition to get here is that a Riemannian manifold, by definition, locally looks like Euclidean space, where the volume of a rectangular box is the product of its side lengths, and we can measure the volume of some particular figure with boxes of any particular orientation and get the same result. We can do this in a manifold too, since the orthogonality of the sides of a box, as well as its side lengths, are all intrinsic. The volume form simply expresses how to do this given any particular chart, by -- for convenience -- using boxes of an orientation that fits with the eigendecomposition of the metric. But the result must be invariant, because the volume of a piece of Euclidean space does not depend on how we turn the coordinate system.

  • $\begingroup$ How do you define an eigenvalue or eigenvector of a bilinear form? $\endgroup$
    – ziggurism
    Oct 9, 2011 at 17:13
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    $\begingroup$ I don't define them for the abstract bilinear form. I'm speaking about eigenvalues and -vectors of the matrix that represents the bilinear form in a particular chart. (It stands to reason that this cannot in itself have a coordinate-free interpretation, because the entire point of the $\sqrt{g}$ factor is to correct for the coordinate-dependent scaling error of $dx^1\wedge\cdots\wedge dx^n$). $\endgroup$ Oct 9, 2011 at 20:44
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    $\begingroup$ This is my point though. Are we really satisfied with a situation where the volume form cannot be defined for the abstract bilinear form? We have to choose a basis, rely on an isomorphism with $\mathbb{R}^n$ and use the canonical metric on $\mathbb{R}^n$ to convert a bilinear form into a linear transformation and then find its eigenvalues. Compare with the symplectic manifold: take the wedge product of the symplectic form with itself $n$ times. Works on any manifold, without choosing a basis or relying on ad hoc features of $\mathbb{R}^n$. $\endgroup$
    – ziggurism
    Nov 30, 2011 at 4:08

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