Ciro Santilli @cirosantilli 37

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 $V$.

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.

In "practice" it is likely "useless", because the functions that it can integrate that Riemann can't are just too funky to appear in practice :-)

Its value is much more indirect and subtle, as in "it serves as a solid basis of quantum mechanics" due to the definition of Hilbert spaces.

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