Adjoints, inverses, transpose

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1-dim'l vector spaces

If $V$ is 1-dim'l, then $V \approx K$, but not canonically.
However, $\End V = K$ canonically: it's which scalar you multiply by
(concretely, $K \to \End V \to K$ (scalars/trace) are inverse isos).
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Transforms of dual bases

Given a basis for $V$, get a dual basis for $V^*$,
  [given w, get ``express $w$ in terms of $v_i$'']
and this transforms by inverse
[enough to check on elementary matrices]:
the map $\GL V \to \GL V^*$ -is- inverse!

Concretely:
  For $V$ a 1-dim'l space, there's a natural identification of
  $V\setminus\set{0} \isoto V^*\setminus\set{0}$, but it's not linear --
  it's $x \mapsto 1/x$.
How?
  Given $v \neq 0 \in V$, let $v^*(w)=\lambda$, where $w=\lambda v$.
[this is a special case of above]

FIXME: this feels like a "classical adjoint"
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That is:
- Consider the functor of (non-deg) $k$-frames,
  $F_k V := V^k \setminus{degenerate ones}$
    [Zariski open, compliment is a determinantal variety]
  (only functorial (for non-deg) on monos)
- Given $T: V -> W$, and $\End V \to \End W$,
  can ask if these maps intertwine/commute
  (with the natural action $V x \End V \to V$)
- Given an iso $V \to W$, there's then an iso
  $\End V \isoto \End W$ (preserving $\GL V \isoto \GL W$),
    by conjugation
  and an iso $F_k V \to F_k W$;
  in particular, an iso $F_n V \to F_n W$.
  [I'm saying this to carefully distinguish what's going on.]

- There's no natural iso $V \to V^*$,
  but adjoint gives an -anti-iso $\End V \isoto \End V^*$
    [product is backwards]
    (preserving $\GL V \to \GL V^*$)
  and there's a -different- -natural- map between frames
  $F_n V \to F_n V^*$ given by ``dual basis''.
It's -obvious- that if you choose a basis (a basepoint of $F_n V$),
then:
  - $\GL V \isoto F_n V$ (as it's a torsor)
  - $V \isoto V^*$ (but from this POV that's kinda weird)
  - the induced map
    $\GL V \to F_n V \to F_n V^* \to \GL V^* \to \GL V$
    is the inverse.
    Concretely, it says ``if I take the standard basis to
    $v_1,\dots,v_n$, give me the map that takes $v_i$ to the standard
    basis'', which is -exactly- what the inverse means.

FIXME: the -point- of this is to understand how the
``classical adjoint'' is really a kind of adjoint/dual.
That is: we have $\GL V \isoto \GL V^*$ -naturally- by adjoint;
  in matrices this is ``transpose'';
  we also have $\GL V \to \GL V$ -naturally- by inverse
  [the above two are anti-isos]
  Composing, we get $\GL V^* = \GL V$ via
  ``adjoint of inverse'', which is an iso of groups.
  That is, $A \mapsto (A^{-1})^*$.
  This intertwines/commutes with $F_n V \to F_n V^*$.

NB: one conclusion: $\End V$ is often better behaved than $V$
  (eg., $\End V^1 = K$ canonically, $\End V = \End V^*$, etc.)
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Transpose/adjoint

NB: $v^*$ is bad notation (b/c has nothing to do with v);
  what's better notation? (maybe v in V, w in V*?)
Given $T \in \End V$, get $T^* \in \End V^*$ via
$T^*(v^*)(w) := v^*(Tw)$.
That is: $T$ acts on vectors, $T^*$ acts on covectors.
In terms of matrices, write $v$ as a column, $v^*$ as a row,
and $T$ as a matrix.
Then $Tv$ is left-multiplication, $T^*v^* = v^*T$ is right
multiplication -- so writing $v^*$ as a column, $T^*$ acts by
left-multiplication by the transpose, recovering
``$T^*$ is the transpose''.
