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The definition may be simple, but it's not very concrete and I'd argue that makes it not strait forward. While examples of vector spaces can be very concrete (t

The definition may be simple, but it's not very concrete and I'd argue that makes it not strait forward. While examples of vector spaces can be very concrete (think R, R^2, R^30), I struggle to think of a concrete example of a multilinear function from vectors and dual vectors in V to numbers in K. On top of that when working with tensors, you don't usually use the definition os a multilinear function at least as far as I remember.

tel 2y ago
Not really to push back as I do agree that this is a bit trickier to get an intuition for than the OP suggests, but the most trivial concrete example of a (1, 1) tensor would just be the evaluation function (v, f) |-> f(v), which, given a metric, corresponds to the inner product.
cshimmin 2y ago
A simple example of a multilinear function is the inner (a.k.a dot) product <a, b>: it takes a vector (b), and a dual vector (a^T), and returns a number. In tensor notation it's typically written δ_ij. It's multilinear because it's linear in each of its arguments separately: <ca, b> = c<a,b> and <a, cb> = c<a,b>. Another simple but less obvious example is a rotation (orthogonal) matrix. It takes a vector as an input, and returns a vector. But a vector itself can be thought of as a linear function that takes a dual vector and returns a number (via the inner product, above!). So, applying the rotation matrix to a vector is a sort of "currying" on the multilinear map, while the matrix alone can be considered a function that takes a vector and a dual vector, and returns a number. In functional notation, you can consider your rotation matrix to be a function (V x V*) -> K, which can in turn be considered a function V -> (V* -> K), where V* is the dual space of V.
- senderista 2y ago
  I think you're describing the evaluation map T(v, w) = w(v), which has type (1,1), rather than the inner product, which has type (2,0). The inner product lets you "raise and lower indices" (i.e. convert between vectors and dual vectors), so you can basically pretend that it is the evaluation map.
- will-burner 2y ago
  The dot product! That's a good example, thank you.
klodolph 2y ago
I think part of this is “if you have a linear algebra background”. There are a few different explanations of tensors, and different explanations make sense for different people.
- cfgauss2718 2y ago
  Yes, the “multi linear map” definition is accessible to an undergraduate who has taken linear algebra. However, the more common meaning of tensor in physics, like the metric tensor of spacetime, requires some more sophisticated background to understand (differential geometry, Lie Groups come to mind).
adrian_b 2y ago
In physics, the first and even now the most important application of multilinear functions, a.k.a. tensors, is in the properties of anisotropic solids. A solid can be anisotropic, i.e. with properties that depend on the direction, either because it is crystalline or because there are certain external influences, like a force or an electric field or a magnetic field that are applied in a certain direction. In (linear) anisotropic solids, a vector property that depends on another vector property is no longer collinear with the source, but it has another direction, so the output vector is a bilinear function of the input vector and of the crystal orientation, i.e. it is obtained by the multiplication with a matrix. This happens for various mechanical, optical, electric or magnetic properties. When there are more complex effects, which connect properties from different domains, like piezoelectricity, which connects electric properties with mechanical properties, then the matrices that describe vector transformations, a.k.a. tensors of the second order, may depend on other such tensors of the second order, so the corresponding dependence is described by a tensor of the fourth order. So the tensors really appear in physics as multilinear functions, which compute the answers to questions like "if I apply a voltage on the electrodes deposited on a crystal in this positions, which will be the direction and magnitude of the displacements of certain parts of the crystal". While in isotropic media you can have relationships between vectors that are described by scalars and relationships between scalars that are also described by scalars, the corresponding relationships for anisotropic media become much more complicated and the simple scalars are replaced everywhere by tensors of various orders. What in an isotropic medium is a simple proportionality becomes a multilinear function in an anisotropic medium. The distinction between vectors and dual vectors appears only when the coordinate system does not use orthogonal axes, which makes all computations much more complicated. The anisotropic solids have become extremely important in modern technology. All the high-performance semiconductor devices are made with anisotropic semiconductor crystals.
cfgauss2718 2y ago
Here’s maybe a useful example. Consider a scalar potential function F on R^3 that describes some nonlinear spring law. At a point p=(x,y,z), the differential dF can be thought of as a (1,0) tensor measuring the spring force. It acts on a particle at p moving with velocity v to give the instantaneous work of the particle on the spring dF(p)(v). Now, suppose that we want to know how this quantity changes when we vary the x coordinate. The x coordinate is also a function of p, we can represent its differential as dx, which is a co-vector(field). The quantity that captures this change can be thought of as a (1,1) tensor field, which is related to the stiffness of the spring potential in the x direction at each point p. In the usual undergraduate setting, this tensor field is given as the hessian of F, call this H. The action of this tensor looks like the product u^T H(p) v, where in our case, u^T = dx(p) = [1 0 0]. A good giveaway for when a “co-vector” appears in a tensor calculation is whenever there is a “row vector” in a matrix operation (most people identify “column” vectors with proper vectors). It’s helpful in this case that “row” rhymes with “co-“.

tel 2y ago

Not really to push back as I do agree that this is a bit trickier to get an intuition for than the OP suggests, but the most trivial concrete example of a (1, 1) tensor would just be the evaluation function (v, f) |-> f(v), which, given a metric, corresponds to the inner product.

cshimmin 2y ago

A simple example of a multilinear function is the inner (a.k.a dot) product <a, b>: it takes a vector (b), and a dual vector (a^T), and returns a number. In tensor notation it's typically written δ_ij. It's multilinear because it's linear in each of its arguments separately: <ca, b> = c<a,b> and <a, cb> = c<a,b>. Another simple but less obvious example is a rotation (orthogonal) matrix. It takes a vector as an input, and returns a vector. But a vector itself can be thought of as a linear function that takes a dual vector and returns a number (via the inner product, above!). So, applying the rotation matrix to a vector is a sort of "currying" on the multilinear map, while the matrix alone can be considered a function that takes a vector and a dual vector, and returns a number. In functional notation, you can consider your rotation matrix to be a function (V x V*) -> K, which can in turn be considered a function V -> (V* -> K), where V* is the dual space of V.

senderista 2y ago
I think you're describing the evaluation map T(v, w) = w(v), which has type (1,1), rather than the inner product, which has type (2,0). The inner product lets you "raise and lower indices" (i.e. convert between vectors and dual vectors), so you can basically pretend that it is the evaluation map.
will-burner 2y ago
The dot product! That's a good example, thank you.

klodolph 2y ago

I think part of this is “if you have a linear algebra background”. There are a few different explanations of tensors, and different explanations make sense for different people.

cfgauss2718 2y ago
Yes, the “multi linear map” definition is accessible to an undergraduate who has taken linear algebra. However, the more common meaning of tensor in physics, like the metric tensor of spacetime, requires some more sophisticated background to understand (differential geometry, Lie Groups come to mind).

adrian_b 2y ago

In physics, the first and even now the most important application of multilinear functions, a.k.a. tensors, is in the properties of anisotropic solids. A solid can be anisotropic, i.e. with properties that depend on the direction, either because it is crystalline or because there are certain external influences, like a force or an electric field or a magnetic field that are applied in a certain direction. In (linear) anisotropic solids, a vector property that depends on another vector property is no longer collinear with the source, but it has another direction, so the output vector is a bilinear function of the input vector and of the crystal orientation, i.e. it is obtained by the multiplication with a matrix. This happens for various mechanical, optical, electric or magnetic properties. When there are more complex effects, which connect properties from different domains, like piezoelectricity, which connects electric properties with mechanical properties, then the matrices that describe vector transformations, a.k.a. tensors of the second order, may depend on other such tensors of the second order, so the corresponding dependence is described by a tensor of the fourth order. So the tensors really appear in physics as multilinear functions, which compute the answers to questions like "if I apply a voltage on the electrodes deposited on a crystal in this positions, which will be the direction and magnitude of the displacements of certain parts of the crystal". While in isotropic media you can have relationships between vectors that are described by scalars and relationships between scalars that are also described by scalars, the corresponding relationships for anisotropic media become much more complicated and the simple scalars are replaced everywhere by tensors of various orders. What in an isotropic medium is a simple proportionality becomes a multilinear function in an anisotropic medium. The distinction between vectors and dual vectors appears only when the coordinate system does not use orthogonal axes, which makes all computations much more complicated. The anisotropic solids have become extremely important in modern technology. All the high-performance semiconductor devices are made with anisotropic semiconductor crystals.

cfgauss2718 2y ago

Here’s maybe a useful example. Consider a scalar potential function F on R^3 that describes some nonlinear spring law. At a point p=(x,y,z), the differential dF can be thought of as a (1,0) tensor measuring the spring force. It acts on a particle at p moving with velocity v to give the instantaneous work of the particle on the spring dF(p)(v). Now, suppose that we want to know how this quantity changes when we vary the x coordinate. The x coordinate is also a function of p, we can represent its differential as dx, which is a co-vector(field). The quantity that captures this change can be thought of as a (1,1) tensor field, which is related to the stiffness of the spring potential in the x direction at each point p. In the usual undergraduate setting, this tensor field is given as the hessian of F, call this H. The action of this tensor looks like the product u^T H(p) v, where in our case, u^T = dx(p) = [1 0 0]. A good giveaway for when a “co-vector” appears in a tensor calculation is whenever there is a “row vector” in a matrix operation (most people identify “column” vectors with proper vectors). It’s helpful in this case that “row” rhymes with “co-“.