# The determinant is the integral of algebra. The integral is the determinant of analysis

This is probably an obvious parallel that most people are aware of, but I only just noticed it the other day and it made me quite excited. The determinant in algebra has a lot in common with the integral in analysis. For example:

1. They are both applied to functions, the integral to integrable functions, the determinant to linear transformations $T:V \rightarrow V$.
2. They are both "sums of products."
3. They both can be used to give a scalar result. (Not always, of course, but this is how they are first developed.)
4. They are both important major structures in algebra and analysis.
5. They are both defined in ways that feel 'backwards'-- the formal definition isn't always useful for calculating them-- then they come to represent multiple important concepts acting as a fulcrum in their fields. (ie. AREA is connected to ANTI-DERIVATIVES... or that SOLUTIONS TO AX=B are connected to LINEAR TRANSFORMATIONS.)
6. They can both be used to give area and volume. (under a curve, or of a parallelepiped)

Question: What mathematical structure encompasses both? (If the answer is category theory, please go slowly with me, I don't understand that stuff yet.)

What else could we add to this list? Are there any problems or proofs that bring these parallels in to the light?

Are there any other mathematical structures that follow the pattern established by these two structures?

Were they developed independently (what I suspect) or is the determinant in some way patterned after the integral or vice versa? (I know my math history and have not come across anything about this.)

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I don't have the time and motivation to give a full answer; I'll just note that there are an awful lot of parallels between integral equations and certain linear algebra problems; e.g. solving a Volterra integral equation through discretization results in the solution of an appropriate lower triangular system, and the Fredholm integral equation corresponds to eigenvalue finding. – J. M. Oct 28 '10 at 12:21
I don't see how this is a good parallel,definite integrals can be understood as linear functionals acting on the space of integrable functions, and thus can be injected into linear algebra. But I find it hard to see how one can generalize the integral and determinant into general idea. IMHO they have many many more differences than similarities (the determinant is distinctly non-linear for instance). – crasic Oct 28 '10 at 12:27
@crasic. Linear operators area decent starting point, but they seem more specialized-- They are in the intersection of integrals and determinants... I'm asking if the union makes any sense. – futurebird Oct 28 '10 at 12:41
+1 crasic. The premise of this question is rather ill-founded. Except 4, which is a tautology, the rest of your "parallels" only hold in a superficial level. – Willie Wong Oct 28 '10 at 12:51
@a little don: here is an obvious difference. The determinant is multiplicative but not additive, and the integral is additive but not multiplicative. – Qiaochu Yuan Oct 28 '10 at 16:19

It's an interesting question. If there were any strong and formalizable analogy it probably would have been developed a long time ago and inscribed in the textbooks. A few observations.

1. The linear algebra analogue of integration is a trace. Determinants are an exponentiated trace. For example $\det \exp A = \exp Tr(A)$ for matrices.

2. If you view integration as solution of differential equations rather than measure, then determinant appears as the Wronskian.

3. Integration (as measure) and determinants are closely related in the theory behind the change of variables formula in integrals: differential forms.

4. The integral is a trace on an infinite-dimensional space (the commutative algebra of functions on, e.g., a closed interval or the real line) while the determinant is specifically finite-dimensional. The Lebesgue measure used in ordinary $n$-dimensional integrals is in some sense defined using determinants (volume), which is why it does not generalize well to infinite dimensional spaces.

5. Thinking about integrals and determinants in terms of formal properties they satisfy leads to "K-theory", specifically $K_1$, but I don't think this produces any deep or striking analogies between the concepts.

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I guess one very natural answer to the question is provided by differential forms. On a smooth $n$-manifold, an $n$-form is a type of "field" whose value at each point is a determinant on the tangent space at that point. (I say "a determinant" because it's only on $\mathbb R^n$ -- or, more generally, an inner product space -- that there's a natural way to define the determinant.) An $n$-form is exactly the type of field that can be integrated over an $n$-manifold in a coordinate-independent way. More generally, a $k$-form on an $n$-manifold is a field that yields a determinant on each $k$-dimensional subspace of each tangent space, and it is the type of field that can be integrated over $k$-dimensional submanifolds. (You can find more about this in my book Introduction to Smooth Manifolds.)

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The determinant is uniquely defined on pretty much anything. Let $V$ be an $n$-dimensional vector space over a field $k$ and let $f:V\to V$ be linear. Then $f$ induces a linear map $\Lambda^n(f):\Lambda^n(V)\to\Lambda^n(V)$, i.e., an element $\Lambda^n(f)\in\mathrm{End}_k(\Lambda^n(V))$. Since $\Lambda^n(V)$ is of dimension one, there is a canonical isomorphism $\mathrm{End}_k(\Lambda^n(V))\cong k$. The image of $\Lambda^n(f)$ in $k$ under this map is precisely the determinant of $f$. – Mariano Suárez-Alvarez Oct 28 '10 at 17:29
@Mariano: I think what Jack Lee is referring to is an identification of Lambda^n(V) with the underlying field, which indeed is not possible to do canonically in general but possible to do canonically if V is an oriented inner product space. (Rather than letting you compute the determinant of a linear map this lets you compute "a" determinant of a tuple of vectors.) – Qiaochu Yuan Oct 28 '10 at 17:34
@Mariano: Sorry, I should have been more explicit. What I was talking about was a determinant as an antisymmetric multilinear scalar-valued function of n vectors. As Qiaochu noted, choosing one such function is equivalent to choosing an identification of $\Lambda^n(V)$ (or $\Lambda^n(V^*)$) with the underlying field. – Jack Lee Oct 28 '10 at 18:50