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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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    @a little don: here is an obvious difference. The determinant is multiplicative but not additive, and the integral is additive but not multiplicative.2010-10-28

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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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    @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.2010-10-28