5

$k$ is a field and $V$ is a finite dimensional $k$ vector space that has an inner product $\langle - , - \rangle$.

If $k = \mathbb{R}$, there is a natural isomorphism $\phi \colon V \rightarrow V^*$, $v \mapsto \langle v, - \rangle$. However, if $k = \mathbb{C}$, $\langle v, - \rangle$ is not linear because $\langle v, cx \rangle = \overline{c} \langle v, x \rangle$. $\langle -, v \rangle$ is linear but in this case $V \rightarrow V^*$, $v \mapsto \langle -, v \rangle$ is not linear.

Are there no natural isomorphism between $V$ and $V^*$ when $k=\mathbb{C}$?

Jendrik Stelzner
  • 16,401
  • Generally, a space and its dual are not naturally isomorphic. A space and the dual of its dual are instead. – xbh Jul 30 '18 at 11:20
  • Maybe $v \mapsto \overline {\langle v, \cdot \rangle}$? 'cause $\overline { \langle v, cx \rangle } = \overline {\overline c \langle v, x \rangle} = c \overline {\langle v,x \rangle }$. – xbh Jul 30 '18 at 11:25
  • 1
    What do you mean by an inner product? – Bob Jul 30 '18 at 14:41
  • i have noticed that physics texts often have $\langle x, y\rangle$ conjugate-linear in $x$ and linear in $y$. – DanielWainfleet Jul 30 '18 at 19:32

1 Answers1

1

Take $\phi: v \mapsto \langle -, v \rangle$. As you've identified, this is not linear but it's almost linear, $\phi(\lambda v +u) = \bar{\lambda}\phi(v) + \phi(u)$.

So instead of an isomorphism between $V^*$ and $V$ we have an isomorphism between $V^*$ and $\bar{V}$, the conjugate space where scalar multiplication is given by $\lambda \cdot v = \bar{\lambda}v$.

Note that the natural isomorphism between $V$ and its double dual still holds.

Also, be careful to remember that an isomorphism between a vector space and its dual only exists in the finite dimensional case (otherwise it's injective but not surjective), and it's not really natural as it relies on a choice of basis (contrast with the isomorphism between $V$ and $(V^*)^*$ which is independent of basis).

Daniel Mroz
  • 2,346
  • 3
    Well given an inner product (as the question states), shouldn't we consider it natural? That's at least how I've seen it. E.g.:

    "When we use the term “canonical isomorphism” we mean that such an isomorphism is defined independently of any choice of bases. ... [I]f $\langle -, - \rangle$ is an inner product on $E$, then Proposition 22.1 shows that the nondegenerate pairing, $\langle -, - \rangle$, induces a canonical isomorphism between $E$ and $E^∗$."

     – A_P Jul 20 '19 at 18:35