Adjoint map and $V \cong V^{**}$

Question

I am confused after following these lines

Note that the adjoint of a bilinear form $B:V\rightarrow V^{*}$ would have the same type $B^{*}:V\rightarrow V^{*}$ (adjoint map) (using $V^{**} \cong V$ ). Only for bilinear forms, we can compare $B$ with $B^{*}$ as they are objects of the same type.

1. What does the word "Type" mean in this context?
2. $V^{**} \cong V$, obviously, this equality is up to isomorphism. Why are we allowed to replace $V^{**}$ as the domain of $B^{*}$ with $V$? Since the objects in $V$ are vectors and objects in $V^{**}$ are maps that take in covectors and spit out scalars. Why is replacing the domain with another set containing elements of different nature does not affect $B^{*}$! Isomorphism preserves the algebraic structure but here we are replacing the objects we take as input with totally different objects. How is this not affecting the map? I cannot come up with an example in this context that clearly shows that this replacement is unimportant.

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The responses made by the community has not yet provided a satisfactory answer to my question.

Answer 1

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1. The term "type" is being employed in a colloquial manner to denote the category of mathematical entity under consideration. The assertion that $B$ and $B^*$ are of the same type signifies that they both function as linear maps, mapping from a vector space $V$ to its dual space $V^*$..

2. The statement $V^{**} \cong V$ asserts the existence of a bijective mapping $I: V \rightarrow V^{**}$ that preserves the vector space structure. This implies that the linear transformation $I$ exhibits the properties of preserving addition and scalar multiplication, specifically, $I(v_1 + v_2) = I(v_1) + I(v_2)$ and $I(c v) = c I(v)$ vectors $\mathbf{\forall} v, v_1, v_2$ in the vector space $V$ and scalar $c$ in the field $\mathbb{R}$ (or any other field under consideration).

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However, the elements of $V$ and $V^{**}$ are fundamentally different objects: $V$ is a set of vectors, while $V^{**}$ is a set of linear functionals on $V^*$. So, when we replace $V^{**}$ with $V$, what we're really doing is replacing each element $v^{**} \in V^{**}$ with its corresponding $v \in V$ under the isomorphism $I$.

It's crucial to note that the "replacement" does not change the nature of the objects. What changes is our perspective, or how we label the objects. The structure-preserving map $I$ allows us to identify each functional in $V^{**}$ with a vector in $V$ in a way that preserves the algebraic structure, and it is in this sense that we say $V^{**} \cong V$. So when we say $B^*$ is a map from $V$ to $V^*$ instead of from $V^{**}$ to $V^*$, what we really mean is that $B^*$ operates on the elements of $V$ that correspond under the isomorphism to the elements of $V^{**}$ that $B^*$ would normally operate on.

For example, consider a finite-dimensional vector space $V$ over the real numbers $\mathbb{R}$. An isomorphism $I: V \rightarrow V^{**}$ might be defined by $I(v)(f) = f(v)$ for all $v \in V$ and $f \in V^*$, where $V^*$ is the set of all linear maps $f: V \rightarrow \mathbb{R}$. Notice that $I(v)$ is indeed a linear map from $V^*$ to $\mathbb{R}$, so it belongs to $V^{**}$. However, by using the isomorphism $I$, we can "identify" each $v \in V$ with its corresponding $I(v) \in V^{**}$, and vice versa.

Answer 2

In the comments you asked

Why do you say "Every statement I make regarding $B^t$, I can make an analogous statement about $\tilde{B}$, and vice-versa"?

Well, the reason I say that is because $\iota$ is an isomorphism. Ok maybe you’re tired of hearing that, but that really is the reason. You also asked what I meant by “information contained in the maps”, well as you’ll see in the examples below this can refer to literally any property of that function (e.g injectivity, surjectivity, bijectivity, rank, nullity, etc). Here are some examples

• $B^t$ and $\tilde{B}$ have the same image. Hence, they both have the same rank. Therefore, $B^t$ is surjective if and only if $\tilde{B}$ is.
• $\ker(B^t)=\iota[\ker \tilde{B}]$. Therefore, $B^t$ is injective if and only if $\tilde{B}$ is.
• putting the previous two statements together, we see that $B^t$ is bijective (hence a linear isomorphism) if and only if $\tilde{B}$ is.
• Fix a $\lambda\in V^*$ and consider the solutions to the equations $\tilde{B}(v)=\lambda$, and $B^t(\zeta)=\lambda$, i.e consider the sets $S_{\tilde{B},\lambda}=\{v\in V\,:\, \tilde{B}(v)=\lambda\}$, and let $S_{B^t,\lambda}=\{\zeta\in V^{**}\,:\, B^t(\zeta)=\lambda\}$. Then, you can immediately conclude that $\iota$ maps $S_{\tilde{B},\lambda}$ bijectively onto $S_{B^t,\lambda}$ (in fact these are affine subspaces, and $\iota$ is an affine bijective mapping between these spaces). In other words, if you know all the solutions to one equation, you know them for the other equation as well.

We can keep going. Ask me any question you like about $B^t$. I can then rephrase that into a question about $\tilde{B}$. Likewise, if you manage to answer one of these questions then by a simple rewording, you also know the answer to the other one. That’s the whole point of $\iota$ being an isomorphism.

So, yes you’re correct that $\tilde{B}$ and $B^t$ are different maps. But so what? They’re related by composition of an isomorphism, so they tell you the same thing (notice also that $\iota$ is not just some random bijection; it is linear which is what gives us for example the statement about kernels in the second bullet point). The stuff I said here goes through for any triple of vector spaces $X,Y,Z$ and linear maps $T:Y\to Z$ and $S:X\to Z$ and a linear isomorphism $\phi:X\to Y$ such that $S=T\circ \phi$. In the case we’re talking about, the reason why we even bother with introducing the $\iota$ is because it is a natural isomorphism; its existence does not rely on me specifying any extra information like a choice of basis. One could (but one doesn’t, without good reason) introduce an arbitrary isomorphism $\phi_1:V\to\Bbb{F}^n$, $\phi_2:V^*\to\Bbb{F}^n$ and $\phi_3:V^{**}\to\Bbb{F}^n$, and talk solely about the corresponding induced linear maps $\Bbb{F}^n\to\Bbb{F}^n$ (or what amounts to the same thing, certain matrix representations). But one doesn’t generally do this because the isomorphisms introduced here are unnatural.