Classical Adjoint

The classical adjoint is (something like)
the induced action on $\Lambda^{n-1} V$.

The fact that you take determinants of $(n-1)$-minors
is a dead giveaway.

How this -actually- works is probably a bit icky;
philosophically, it should be something like:
``$\Lambda^1 V + \Lambda^{n-1} V = \Lambda^n V$
  so when you multiply 'em together, you just get
  `multiplication by the determinant'

Concretely, an inner product for $V$ yields an isomorphism
$V = \Lambda^1 V \isoto \Lambda^{n-1} V$
(this is basic Hodge theory)
via $e_1 \mapsto e_2 \wedge \dots \wedge e_n$
and $e_k \mapsto
e_1 \wedge \dots \wedge \hat e_k \wedge \dots \wedge e_n
= \bigwedge_{i\neq k} e_i$
and so forth.
(that's why there's the classical adjoint of a -matrix-,
 not an -operator-; without the basis, which yields the inner product,
 you get an operator on $\Lambda^{n-1} V$, not on $V$.)

[and this starts to explain why the adjoint works;
 it's something like ``the action on the perp'',
 which combines with the action on the vector
 to yield the action on the whole space,
 namely the determinant]

[The slightly mysterious thing is:
 you then -multiply- these as operators on $V$,
 and then get a scalar by the determinant -- which is a bit weird,
 since we think of the determinant as scaling $\Lambda^n V$,
 not $V$.
 Perhaps there's some funky duality going on here;
 $\End \Lambda^n V \to \End \Lambda^{n-1} V \approx \End V$??]

Incidentally, there's a few lil' properties of the adjoint
(that I learned from that Jacob book for 196)
that make a lot of sense from this POV
and that generalize nicely.

EG, if $T$ has rank $n-1$, then $\adj T$ has rank $1$;
if $T$ has rank $\leq n-2$, then $\adj T = 0$
More generally:
- if $T$ has rank $< k$, then $\Lambda^k T = 0$
  This generalizes the known cases
   $k=1,n-1,n$; this first is obvious, and the last
   is the determinant.
  It's geometrically clear from the action on $k$-planes
    (thought of as represented by $k$-fold wedges);
    if $T$ has rank $<k$, it sends a $k$-plane into something smaller,
    so it kills $k$th wedges, so $\Lambda^k T = 0$.

[oh, and the other statement should generalize to:
 if $T$ has rank $k$, then $\Lambda^k T$ has rank $1$
 ...and more properly you can perhaps write the ranks
 of $\Lambda^i T$ in terms of $i, \rk T$;
 maybe it's $\binom{\rk T}{i}$?]

remember:
- to understand action on $k$-planes,
  look at $\Lambda^k T$
- $k$-minors -means- $\Lambda^k T$

BTW, this all shows that the rank filtration on $\End V$ arises from
the simple $(0,V)$ (zero vector, whole space) filtration on
$\Lambda^k V$, pulled back functorially:
you're rank $<k$ iff you vanish in $\End \Lambda^k V$

(Presumably the other Schur functors aren't so interesting
 algebraically? Or are $\Sym$ and $\Alt$ just the ``easy'' cases,
 and there's a cool general theory?
 Also, for general tensor powers you want to talk about
 non-commutative geometry.)

BTW, can we relate this back to thinking of $\End V = V^* \otimes V$,
and thinking of rank as ``minimal number of summands''?
(in some sense, sure: you get the exterior powers
 by combining summands in some way)
