Ciro Santilli @cirosantilli  40

Given a linear operator $A$ over a space $S$ that has a inner product defined, we define the adjoint operator $A^{†}$ (the $†$ symbol is called "dagger") as the unique operator that satisfies:

For the typical case of a linear form over ${\mathbb{R}}^n$, the form can be seen just as a row vector with n elements, the full form being specified by the value of each of the basis vectors.

The dual space of a vector space $V$, sometimes denoted $V^{\ast }$, is the vector space of all linear forms over $V$ with the obvious addition and scalar multiplication operations defined.

Since a linear form is completely determined by how it acts on a basis, and since for each basis element it is specified by a scalar, at least in finite dimension, the dimension of the dual space is the same as the $V$, and so they are isomorphic because all vector spaces of the same dimension on a given field are isomorphic, and so the dual is quite a boring concept in the context of finite dimension.

Infinite dimension seems more interesting however, see: en.wikipedia.org/w/index.php?title=Dual_space&oldid=1046421278#Infinite-dimensional_case

One place where duals are different from the non-duals however is when dealing with tensors, because they transform differently than vectors from the base space $V$.

This does not seem to go deep into the Standard Model as Physics from Symmetry by Jakob Schwichtenberg (2015), appears to focus more on more basic applications.

But because it is more basic, it does explain some things quite well.

The author seems to have uploaded the entire book by chapters at: www.physics.drexel.edu/~bob/LieGroups.html

And the author is the cutest: www.physics.drexel.edu/~bob/Personal.html.

Overview:

Elements of a Lie algebra can (should!) be seen a continuous analogue to the generating set of a group in finite groups.

For continuous groups however, we can't have a finite generating set in the strict sense, as a finite set won't ever cover every possible point.

But the generator of a Lie algebra can be finite.

And just like in finite groups, where you can specify the full group by specifying only the relationships between generating elements, in the Lie algebra you can almost specify the full group by specifying the relationships between the elements of a generator of the Lie algebra.

This "specification of a relation" is done by defining the Lie bracket.

The reason why the algebra works out well for continuous stuff is that by definition an algebra over a field is a vector space with some extra structure, and we know very well how to make infinitesimal elements in a vector space: just multiply its vectors by a constant $c$ that cana be arbitrarily small.

Every Lie algebra corresponds to a single simply connected Lie group.

The Baker-Campbell-Hausdorff formula basically defines how to map an algebra to the group.

Bibliography:

Most commonly refers to: exponential map.

Solution $Z$ for given $X$ and $Y$ of:where $e$ is the exponential map.

If we consider just real number, $Z=X+Y$, but when X and Y are non-commutative, things are not so simple.

Furthermore, TODO confirm it is possible that a solution does not exist at all if $X$ and $Y$ aren't sufficiently small.

This formula is likely the basis for the Lie group-Lie algebra correspondence. With it, we express the actual group operation in terms of the Lie algebra operations.

Notably, remember that a algebra over a field is just a vector space with one extra product operation defined.

Vector spaces are simple because all vector spaces of the same dimension on a given field are isomorphic, so besides the dimension, once we define a Lie bracket, we also define the corresponding Lie group.

Since a group is basically defined by what the group operation does to two arbitrary elements, once we have that defined via the Baker-Campbell-Hausdorff formula, we are basically done defining the group in terms of the algebra.

Cardinality $\le$ dimension of the vector space.