I Definition of the special unitary group

[jbergman]

Path connectedness?

For finite dimensional groups $U(n)$ is path connected (I think). I'm not familiar enough with the infinite dimensional case to comment.

 

[redtree]

I will ask again, can you give a matrix free definition of $SL_2$? So that we know what you expect as an answer.

For a finite dimensional, matrix-free formulation of $GL(n,\mathbb{R})$ with a tensor determinate operation similar to that of a matrix (https://hal.science/hal-03298805v1/file/tensors.pdf):

$$GL(n,\mathbb{R}) = \{ T_{i j} \in GL(n,\mathbb{R}), \dim{i}, \dim{j} = n: \det{T_{i j}} \neq 0 \}$$

such that one might define the finite-dimensional, matrix-free $SL(n,\mathbb{R})$ formulation:

$$SL(n,\mathbb{R}) = \{ T_{i j} \in SL(n,\mathbb{R}), \dim{i}, \dim{j} = n: \det{T_{i j}} =1 \}$$

 

[martinbn]

For a finite dimensional, matrix-free formulation of $GL(n,\mathbb{R})$ with a tensor determinate operation similar to that of a matrix (https://hal.science/hal-03298805v1/file/tensors.pdf):

$$GL(n,\mathbb{R}) = \{ T_{i j} \in GL(n,\mathbb{R}), \dim{i}, \dim{j} = n: \det{T_{i j}} \neq 0 \}$$

such that one might define the finite-dimensional, matrix-free $SL(n,\mathbb{R})$ formulation:

$$SL(n,\mathbb{R}) = \{ T_{i j} \in SL(n,\mathbb{R}), \dim{i}, \dim{j} = n: \det{T_{i j}} =1 \}$$

!!!

 

[redtree]

!!!

What were you hoping to see?

 

[martinbn]

What were you hoping to see?

What you've written uses matrices, and you wanted a definition without them. More importantly it is written in a way that makes no sense and is just a tautology.

 

[Euge]

TL;DR Summary: Defining the special unitary group independent of matrix representation

In matrix representation, the special unitary group is distinguished from the more general unitary group by the sign of the matrix determinant. However, this presupposes that the special unitary group is formulated in matrix representation. For a unitary group action NOT formulated in matrix representation, what differentiates a special unitary group action from a more general unitary group action, such as in the infinite dimensional case?

If $H$ is a separable Hilbert space and $T$ is a trace-class operator on $H$ (i.e., a bounded linear operator on $H$ whose sequence of singular values is summable), then one may define the determinant of $I + T$ as the product $\prod_{j = 1}^\infty (1 + \lambda_j(T))$ where the $\lambda_j(T)$ are the eigenvalues of $T$. The space $B_1(H)$ of trace-class operators on $H$ is an ideal of the space $B(H)$ of bounded linear operators on $H$. So if $U(H)$ is the group of all unitary operators on $H$, it would make sense to define $SU(H)$ as the set of all $\Lambda \in U(H)$ such that $\Lambda - I \in B_1(H)$ and $\det \Lambda = 1$. This will be a subgroup of $U(H)$, and when $H = \mathbb{C}^n$, $U(H)$ and $SU(H)$ are canonically identified with the usual unitary matrix groups $U(n)$ and $SU(n)$, respectively.

 

[redtree]

$$GL(n,\mathbb{R}) = \{ T^{i j} \in V^i \otimes W^j, \dim{i}, \dim{j} = n: \det{T^{i j}} \neq 0 \}$$

$$SL(n,\mathbb{R}) = \{ T^{i j} \in V^i \otimes W^j, \dim{i}, \dim{j} = n: \det{T^{i j}} =1 \}$$

 

[mathwonk]

I am a bit puzzled by most of this discussion, since in finite dimensions neither the hermitian property nor the determinant one property are dependent on matrices. A hermitian linear map on a vector space V is one that preserves a hermitian inner product, and if V is finite dimensional, say dim(V) = n, then its nth wedge product V^n, is one dimensional, and thus the induced map on it is canonically given by a scalar, the determinant. The only issue for me is to define a determinant in infinite dimensions and Euge has addressed this. Of course even in finite dimensions, all unitary groups associated to vector spaces equipped with Hermitian forms, are isomorphic to the standard ones defined by matrices, so it is understandable to assume that matrices are natural to their definition.

 

[martinbn]

I am a bit puzzled by most of this discussion, since in finite dimensions neither the hermitian property nor the determinant one property are dependent on matrices. A hermitian linear map on a vector space V is one that preserves a hermitian inner product, and if V is finite dimensional, say dim(V) = n, then its nth wedge product V^n, is one dimensional, and thus the induced map on it is canonically given by a scalar, the determinant. The only issue for me is to define a determinant in infinite dimensions and Euge has addressed this. Of course even in finite dimensions, all unitary groups associated to vector spaces equipped with Hermitian forms, are isomorphic to the standard ones defined by matrices, so it is understandable to assume that matrices are natural to their definition.

You can avoid the word matrix this way, but the unitary group is still defined as a subgroup of the general linear group i.e. through a representation. The way I understood the question, he wants a definition that does not involve representations of the group. He wants to seperate the group and its representations. As an analogy every finite group is a subgroup of a permutation group. But some specific groups are defined without the use of a permutation group. How can it be done say for the alternating group?

 

[mathwonk]

If V is a vector space, then the "general linear group" associated to it , GL(V), is the group of invertible linear transformations of V. Until a basis is chosen this is not a matrix group in my view. There is no (matrix) representation here. Similarly given a complex vector space V and a hermitian form f, the unitary group of the pair (V,f) is the subgroup of GL(V) preserving f. Again, until a basis is chosen, there is no representation as matrices.

I would also say, that not all finite groups are subgroups of a (finite) permutation group. Rather, they are all isomorphic (in many different ways) to such subgroups.

 

[jbergman]

If V is a vector space, then the "general linear group" associated to it , GL(V), is the group of invertible linear transformations of V. Until a basis is chosen this is not a matrix group in my view. There is no representation here. Similarly given a complex vector space V and a hermitian form f, the unitary group of the pair (V,f) is the subgroup of GL(V) preserving f. Again, until a basis is chosen, there is no representation as matrices.

I would also say, that not all finite groups are subgroups of a (finite) permutation group. Rather, they are all isomorphic (in many different ways) to such subgroups.

Given $L \in U(V,f)$ we have it's canonical action on the nth exterior power as $$L(v_1\wedge ... \wedge v_n) = L(v_1) \wedge ... \wedge L(v_n) $$

However, don't you have to choose a basis for $V$ to assert that a particular top power element $v_1\wedge ... \wedge v_n$ is the basis for that space to determine which $L$ are in the special unitary group?

 

[jbergman]

Given $L \in U(V,f)$ we have it's canonical action on the nth exterior power as $$L(v_1\wedge ... \wedge v_n) = L(v_1) \wedge ... \wedge L(v_n) $$

However, don't you have to choose a basis for $V$ to assert that a particular top power element $v_1\wedge ... \wedge v_n$ is the basis for that space to determine which $L$ are in the special unitary group?

Actually nevermind, I guess the claim is that
$$L(v_1\wedge ... \wedge v_n) = L(v_1) \wedge ... \wedge L(v_n) = v_1\wedge ... \wedge v_n $$

Is the condition we want for special groups.

 

[mathwonk]

yes, this is the very interesting fact that any linear self map of a one dimensional space is multiplication by a number. So we only need that V^n is one dimensional. e.g. a real number is given by an ordered pair of segments on a line. Thus choosing coordinates on a line is usually done by choosing an origin and a unit point. Then any other point, together with the origin, gives a second segment, which when compared to the first unit segment, gives a number.
Given any one dimensional vector space, we have an origin, the zero vector. Then any other two vectors, in order, give two ordered segments, hence a number. or more simply, given any two ≠0 vectors v,w in a one dimensional space, there is a unique number c such that w = cv.

On a Riemann surface, this yields the phenomenon that a meromorphic function Is defined by (the quotient of) two sections of a line bundle. I.e. given s,t, non zero sections of a line bundle, at every point p, their values s(p), t(p), both lie in the same line, hence their ratio is a number, without ever choosing a basis for the one dimensional linear fiber. I.e. even though neither s(p) nor t(p) is a number, still s(p)/t(p) is a number. so s/t is a meromorphic function.

Similarly, in calculus, given a smooth function y(x), we have the objects dy and dx, each of which is an element of the cotangent bundle to the line, hence at each point p, dy(p) and dx(p) are elements of the same one dimensional cotangent space; thus although neither dy(p) nor dx(p) is a number, for each p, the ratio dy(p)/dx(p) is a number; i.e. dy and dx are sections of the same line bundle, called the "canonical bundle" when discussing Riemann surfaces, and hence dy/dx is a function.

As for finite groups, imagine a finite group of isometries of the sphere. How is that to be viewed as a subgroup of a finite permutation group? It is possible of course, but in several ways.

Or just imagine the group of rotations of a cube. To represent it as a finite permutation group, one must choose a finite set on which it acts faithfully. Of course there are several, but none are distinguished. It acts on the vertices, the edges, the faces, the diagonals, the axes, indeed on the orbit of any point at all. These various choices give homomorphisms to the permutation groups S(8), S(12), S(6), S(4), S(3), S(24) if the point is general enough.
One may want to use as small a set as possible, e.g. the 3 axes, but this representation is not faithful, since the rotation group has order 24 and that permutation group has order 6. Although the vertices may seem the most natural, the most efficient choice is thus the 4 diagonals, which gives an isomorphism onto the full permutation group S(4) of order 24, (although to me the fact this is an isomorphism is not exactly obvious; I had to re-read my argument in my own class notes).

 

[martinbn]

If V is a vector space, then the "general linear group" associated to it , GL(V), is the group of invertible linear transformations of V. Until a basis is chosen this is not a matrix group in my view. There is no (matrix) representation here. Similarly given a complex vector space V and a hermitian form f, the unitary group of the pair (V,f) is the subgroup of GL(V) preserving f. Again, until a basis is chosen, there is no representation as matrices.

By representation i meant a homomorphism to a general linear group. The unitary group is defined as a subgroup of the general linear group. So there is a representation. And the way i understood the question is that yhe OP wants a definition of thw group which is seperate from the representation theory of the group.

So i dont think the way you suggest to avoid matrices in the definition would satisfy the OP. I might be wrong, he will clarify.

I would also say, that not all finite groups are subgroups of a (finite) permutation group. Rather, they are all isomorphic (in many different ways) to such subgroups.

Yes, of course.

 

[mathwonk]

by the way, it seems alternating groups An, for n ≥ 9 are completely characterized by their abstract group structure, namely by the facts that they are non abelian, simple, and of order n!/2.
https://en.wikipedia.org/wiki/List_of_finite_simple_groups

 

[jbergman]

By representation i meant a homomorphism to a general linear group. The unitary group is defined as a subgroup of the general linear group. So there is a representation. And the way i understood the question is that yhe OP wants a definition of thw group which is seperate from the representation theory of the group.

So i dont think the way you suggest to avoid matrices in the definition would satisfy the OP. I might be wrong, he will clarify.

Yes, of course.

I took it slightly differently since @redtree mentioned matrices in his original post. I think he might jave been using the word representation somewhat imprecisely. The distinction, I would make is that @mathwonk has presented a way to characterize the special unitary groups for $(V, f)$ without having chosen a basis.

 

[martinbn]

I took it slightly differently since @redtree mentioned matrices in his original post. I think he might jave been using the word representation somewhat imprecisely. The distinction, I would make is that @mathwonk has presented a way to characterize the special unitary groups for $(V, f)$ without having chosen a basis.

May be, I expected that he would say something if he is still around.