Also, I’m curious what you think the connection is between the “L2 is connected to bilinear forms” and “L2 is the only Lp metric invariant under nontrivial change of basis”, if it’s easy to state.
This was what I was trying to vaguely gesture towards with the derivation of the “transpose = inverse” characterization of L2-preserving matrices; the idea was that the argument was a natural sort of thing to try, so if it works to get us a characterization of the Lp-preserving matrices for exactly one value of p, then that’s probably the one that has a different space of Lp-preserving matrices than the rest. But perhaps this is too sketchy and mysterian. Let’s try a dimension-counting argument.
Linear transformations Rn→Rn and bilinear forms Rn×Rn→R can both be represented with n×n matrices. Linear transformations act on the space of bilinear forms by applying the linear transformation to both inputs before plugging them into the bilinear form. If the matrix A represents a linear transformation and the matrix B represents a bilinear form, then the matrix representing the bilinear form you get from this action is ATBA. But whatever, the point is, so far we have an n2-dimensional group acting on an n2-dimensional space. But quadratic forms (like the square of the L2 norm) can be represented by symmetricn×n matrices, the space of which is (n+12)-dimensional, and if B is symmetric, then so is ATBA. So now we have an n2-dimensional group acting on a (n+12)-dimensional space, so the stabilizer of any given element must be at least n2−(n+12)=(n2) dimensional. As it turns out, this is exactly the dimensionality of the space of orthogonal matrices, but the important thing is that this is nonzero, which explains why the space of orthogonal matrices must not be discrete.
Now let’s see what happens if we try to adapt this argument to Lp and p-linear forms for some p≠2.
With p=1, a linear transformation preserving a linear functional corresponds to a matrix A preserving a row vector φ in the sense that φA=φ. You can do a dimension-counting argument and find that there are tons of these matrices for any given row vector, but it doesn’t do you any good because 1 isn’t even so preserving the linear functional doesn’t mean you preserve L1 norm.
Let’s try p=4, then. A 4-linear form Rn×Rn×Rn×Rn→R can be represented by an n×n×n×n hypermatrix, the space of which is n4-dimensional. Again, we can restrict attention to the symmetric ones, which are preserved by the action of linear maps. But the space of symmetric n×n×n×n hypermatrices is (n+34)-dimensional, still much more than n2. This means that our linear maps can use up all of their degrees of freedom moving a symmetric 4-linear form around to different 4-linear forms without even getting close to filling up the whole space, and never gets forced to use its surplus degrees of freedom with linear maps that stabilize a 4-linear form, so it doesn’t give us linear maps stabilizing L4 norm.
This was what I was trying to vaguely gesture towards with the derivation of the “transpose = inverse” characterization of L2-preserving matrices; the idea was that the argument was a natural sort of thing to try, so if it works to get us a characterization of the Lp-preserving matrices for exactly one value of p, then that’s probably the one that has a different space of Lp-preserving matrices than the rest. But perhaps this is too sketchy and mysterian. Let’s try a dimension-counting argument.
Linear transformations Rn→Rn and bilinear forms Rn×Rn→R can both be represented with n×n matrices. Linear transformations act on the space of bilinear forms by applying the linear transformation to both inputs before plugging them into the bilinear form. If the matrix A represents a linear transformation and the matrix B represents a bilinear form, then the matrix representing the bilinear form you get from this action is ATBA. But whatever, the point is, so far we have an n2-dimensional group acting on an n2-dimensional space. But quadratic forms (like the square of the L2 norm) can be represented by symmetric n×n matrices, the space of which is (n+12)-dimensional, and if B is symmetric, then so is ATBA. So now we have an n2-dimensional group acting on a (n+12)-dimensional space, so the stabilizer of any given element must be at least n2−(n+12)=(n2) dimensional. As it turns out, this is exactly the dimensionality of the space of orthogonal matrices, but the important thing is that this is nonzero, which explains why the space of orthogonal matrices must not be discrete.
Now let’s see what happens if we try to adapt this argument to Lp and p-linear forms for some p≠2.
With p=1, a linear transformation preserving a linear functional corresponds to a matrix A preserving a row vector φ in the sense that φA=φ. You can do a dimension-counting argument and find that there are tons of these matrices for any given row vector, but it doesn’t do you any good because 1 isn’t even so preserving the linear functional doesn’t mean you preserve L1 norm.
Let’s try p=4, then. A 4-linear form Rn×Rn×Rn×Rn→R can be represented by an n×n×n×n hypermatrix, the space of which is n4-dimensional. Again, we can restrict attention to the symmetric ones, which are preserved by the action of linear maps. But the space of symmetric n×n×n×n hypermatrices is (n+34)-dimensional, still much more than n2. This means that our linear maps can use up all of their degrees of freedom moving a symmetric 4-linear form around to different 4-linear forms without even getting close to filling up the whole space, and never gets forced to use its surplus degrees of freedom with linear maps that stabilize a 4-linear form, so it doesn’t give us linear maps stabilizing L4 norm.