# Natural isomorphisms

1. Jan 11, 2009

### jojo12345

The text I'm reading explains how there is a natural isomorphism between a vector space and the dual of the dual of the vector space. The author explains that this is so because the isomorphism he defines makes no reference to a specific basis of the vector space. I understand that natural isomorphisms fall under the umbrella of category theory. Why are natural isomorphisms significant?

2. Jan 11, 2009

### Hurkyl

Staff Emeritus
The direct meaning of 'natural isomorphism' in this case is that for each vector space V, you have an isomorphism $\phi_V : V \to V^{**}$, and furthermore, for every linear transformation $T : V \to W$, you have the identity $\phi_W(T(v)) = T^{**}(\phi_V(v))$.

($T^{**}$ is the corresponding action of T on the double-dual)

If you're comfortable with depicting linear tranformations as diagrams, you have

$$\begin{array}{rcl} V &\xrightarrow{\phi_V}& V^{**} \\ T \downarrow & & \downarrow T^{**} \\ W &\xrightarrow{\phi_W}& W^{**}$$

The practical effect is that the naturality condition of a transformation ensures that you generally don't have to worry about when it applies, since you can do (certain) things in any order you please.

Compare with linear transformations on vector spaces; the linear transformation "commutes" with addition and scalar multiplication, so if you wanted to compute something like $T(rv + w)$, you can either do the operations first then apply T, or apply T then do the operations, or some combination of the two.

In fact, a natural transformation is defined as a homomorphism of "constructions". (More precisely, of functors) In this case, it is to the "double dual" functor from the the "identity" construction that leaves a vector space / linear transformation unchanged.

If you're familiar with homotopies, natural isomorphisms are related. (Both in a superficial and in a meaningful sense)

Sometimes, though, natural is really just used in the plain-English sense. Many common examples of natural isomorphisms in the technical sense turn out to be natural in the plain-English sense too. (like the double dual)

Last edited: Jan 11, 2009
3. Jan 11, 2009

### jojo12345

how do you define the corresponding action of T on the double-dual?

4. Jan 12, 2009

### Hurkyl

Staff Emeritus
By applying the dual twice, just as for the vector spaces themselves.

If you haven't seen it, $T^*$ is a map in the reverse direction, $W^* \to V^*$, and is given as follows:
$$T^*(\phi)(v) = \phi(T(v))[/itex] where $\phi \in W^*$ and $v \in V$. More succinctly, $T^*\phi = \phi \circ T$. (remember that $T^*(w)$ is a linear functional on V, so it can be evaluated at v) 5. Jan 12, 2009 ### jojo12345 so let me see if I'm interpreting the corresponding action of [tex]T$$ on the double dual correctly:

$$(T^{**}\phi_{V}(v))(w)=\phi_{V}(v)\circ{}T^{*}(w)=\phi_{V}(v)(w\circ T)$$

where $$v\in{V}$$ and $$w\in{W^{*}}$$?

What is the special feature of the isomorphisms $$\phi_{V}$$ that leads to the identity you mentioned?

6. Jan 12, 2009

### jojo12345

I actually read more about category theory and I think I can see how to start to prove this identity. Right now I'm trying to prove that the map F from the category of vector spaces to the category of vector spaces defined by : $$F(V)=V^{**},F(T)=T^{**}$$ ,where $$V$$is a vector space and $$T$$ is a linear map $$T:V\rightarrow W$$ with W a vector space, is a functor. I'll post when/if I figure it out.

7. Jan 12, 2009

### Hurkyl

Staff Emeritus
Right.

8. Jan 13, 2009

### jojo12345

I've managed to prove $$F$$ is a functor by proving the function $$f:vec\rightarrow vec$$ ,defined by $$f(V)=V^{*},f(T)=T^{*}$$, is a contravariant functor and that the composition of two contravariant functors is a functor.

Now to prove that identity. let $$F$$ be the identity functor on the category of finite dimensional vector spaces and $$G$$ be the functor that sends vector spaces to their double duals. Define the following function from a vector space X to its double-dual: $$\phi_{X}:F(X)\rightarrow G(X)$$ as $$\phi_{X}(x)=\bar{x}$$ where $$\bar{x}(f)=f(x)\forall f\in X^{*}$$. This map $$\phi_{X}$$ is linear and invertible. I can prove this, but I won't here.

Now consider any vector space homomorphism between vector spaces X and Y, $$T:X\rightarrow Y$$.

\begin{align*} G(T)\circ\phi_{X}=T^{**}\circ\phi_{X}&\Rightarrow \exists z\in{Y}, G(T)\circ\phi_{X}(x)=\phi_{Y}(z)\\ {}&\Rightarrow z=\phi^{-1}_{Y}\circ T^{**}\circ\phi_{X}(x)\\ {}&\Rightarrow z=\phi^{-1}_{Y}(T^{**}\bar{x})=\phi^{-1}_{Y}(\bar{x}\circ T^{*})=\phi^{-1}_{Y}(\bar{Tx})=Tx=F(T)(x)\\ \end{align*}

The last line is true because $$\forall y\in Y^{*}, \bar{Tx}(y)=y\circ T(x)=T^{*}y(x)=\bar{x}\circ T^{*}(y)$$

I think that's it.

9. Jan 14, 2009

### jojo12345

Is it reasonable to assume that if a vector space isomorphism doesn't depend on the components of vectors involved, then it is natural? I can't think of an exact way to formulate the question...