Consider the polynomial algebra $\mathbb{C}[x,y]$ with operation $\Delta(x)=x\otimes x- y\otimes y $ and $\Delta(y)=x\otimes y+ y\otimes x$. Find the operations counit $\epsilon$ and antipode $S$ such that $\mathbb{C}[x,y]$ with the $\Delta$ will become the Hopf algebra.
My solution:
Step 1.
$x=x\epsilon(x)- y\epsilon (y)$ and $y=x\epsilon(y)+y\epsilon (x)$. It implies that $\epsilon (x)=1$ and $\epsilon (y)=0.$
Step 2.
$\epsilon(x)=xS(x)- yS(y)$ and $\epsilon(y)=xS(y)+yS(x)$. This system defines the values of $S$, but it seems that they aren't polynomials. What wrong with my solution?
I think your calculations are correct. It follows from the condition $$ x = \sum_{(x)} x_{(1)} \varepsilon( x_{(2)} ) = x \varepsilon(x) - y \varepsilon(y) $$ that $\varepsilon(x) = 1$, and then also $\varepsilon(y) = 0$. The antipode does then have to satisfy $$ 1 = \varepsilon(x) = \sum_{(x)} x_{(1)} S( x_{(2)} ) = x S(x) - y S(y)\,. $$ But the right hand side of this equation has no constant part. This equality is thus not possible.
I think there are two ways of fixing this example.
First way
Let $T$ be a two-dimensional vector space with basis $c$, $s$. One can make $T$ into a coalgebra by setting $$ \Delta(c) = c \otimes c - s \otimes s \, \quad \Delta(s) = c \otimes s + s \otimes c \,. $$ The counit is given by $$ \varepsilon(c) = 1 \,, \quad \varepsilon(s) = 0 \,. $$ This is (sometimes) called the trigonometric coalgebra. One should think about the basis elements $c$ and $s$ as the functions $\cos$ and $\sin$, and the comultiplication as a version of the addition theorems for cosine and sine.
This heuristic becomes justified by the following example.
Second way, preparation
Let $X$ be a set and let $\mathbb{k}$ be a field. Let $F(X)$ be the set of functions from $X$ to $\mathbb{k}$. This is a algebra via the pointwise addition, scalar multiplication and multiplication of functions, i.e. via \begin{align*} (f_1 + f_2)(x) &= f_1(x) + f_2(x) \, \\ (\lambda f)(x) &= \lambda f(x) \,, \\ (f_1 f_2)(x) &= f_1(x) f_2(x) \,. \end{align*}
We have a map \begin{align*} \Phi \colon F(X) \otimes F(X) &\to F(X \times X) \,, \\ f_1 \otimes f_2 &\mapsto [ (x_1, x_2) \mapsto f_1(x_1) f_2(x_2) ] \,. \end{align*} This map is an injective homomorphism of algebras. If the set $X$ is finite then by comparing dimensions we find that the homomorphism $\Phi$ is an isomorphism.
Let now $G$ be a finite group. The multiplication of $G$ is a map $$ m \colon G \times G \to G \,. $$ This map induces a homomorphism of algebras $$ m^* \colon F(G) \to F(G \times G) \,, \quad f \mapsto f \circ m \,. $$ More explicitely, $$ m^*(f)(g_1, g_2) = f(g_1 g_2) \,. $$ Under the isomorphism $\Phi^{-1}$ we can regard the homomorphism $m^*$ as a homomorphism of algebras $$ \Delta \colon F(G) \to F(G) \otimes F(G) \,. $$ More explicitely, in terms of Sweedler notation, $$ f( g_1 g_2 ) = \sum_{(f)} f_{(1)}( g_1 ) f_{(2)}( g_2 ) \,. $$
Let us denote the unit element of $G$ by $e$. The inclusion map $\{ e \} \to G$ induces an algebra homomorphism $F(G) \to F(\{e\})$. We may identify $F(\{e\})$ with the ground field $\mathbb{k}$ via the evaluation map $f \mapsto f(e)$. We then get overall a homomorphism of algebras $$ \varepsilon \colon F(G) \to \mathbb{k} \,, \quad f \mapsto f(e) \,. $$
Lastly, the inversion map $$ i \colon G \to G \,, \quad g \mapsto g^{-1} $$ induces an algebra homomorphism $$ S := i^* \colon F(G) \to F(G) \,, \quad f \mapsto f \circ i \,. $$ More explicitely, $$ S(f)(g) = f( g^{-1} ). $$
The algebra $F(G)$ together with the homomorphisms $\Delta$, $\varepsilon$ and $S$ is a Hopf algebra. For an inifinite group $G$ this construction does in general not work because the injective homomorphism $\Phi$ won’t be an isomorphism.
Second way
We would now like to take $\mathbb{k} = \mathbb{R}$ and for the finite group $G$ the additive group $(\mathbb{R}, +)$. But this group isn’t finite. We can try to circumvent this problem by considering instead of $F(\mathbb{R})$ only a suitable subalgebra $H$, such that $m^*(H)$ is contained in $\Phi( H \otimes H )$, and $S(H) \subseteq H$. We can then make $H$ into a Hopf algebra in the same way as above.
For our example we choose $H$ as the subalgebra of $F(\mathbb{R})$ which is generated by the two functions $\cos$ and $\sin$. We have \begin{align*} m^*(\cos)(x, y) &= \cos(x + y) \\ &= \cos(x) \cos(y) - \sin(x) \sin(y) \\ &= \Phi( \cos \otimes \cos - \sin \otimes \sin )(x, y) \end{align*} and \begin{align*} m^*(\sin)(x,y) &= \sin(x + y) \\ &= \cos(x)\sin(y) + \sin(x)\cos(y) \\ &= \Phi( \cos \otimes \sin + \sin \otimes \cos )(x,y) \,, \end{align*} as well as $$ S(\cos)(x) = \cos(-x) = \cos(x) $$ and $$ S(\sin)(x) = \sin(-x) = -\sin(x) \,. $$
We now find that the algebra $H$ is a Hopf algebra with comultiplication \begin{align*} \Delta(\cos) &= \cos \otimes \cos - \sin \otimes \sin \,, \\ \Delta(\sin) &= \cos \otimes \sin + \sin \otimes \cos \,, \end{align*} counit $$ \varepsilon( \cos ) = \cos(0) = 1 \,, \quad \varepsilon( \sin ) = \sin(0) = 0 \,, $$ and antipode $$ S(\cos) = \cos \,, \quad S(\sin) = -\sin \,. $$
The functions $\sin$ and $\cos$ satisfy the relation $$ \cos^2 + \sin^2 = 1 \,. $$ We can follow from this that the algebra $H$ is isomorphic to the algebra $$ H' = \mathbb{R}[c, s] / (c^2 + s^2 - 1) \,, $$ such that $c$ corresponds to $\cos$ and $s$ corresponds to $s$. The algebra $H'$ does now become a Hopf algebra with comultiplication $$ \Delta(c) = c \otimes c - s \otimes s \,, \quad \Delta(s) = c \otimes s - s \otimes c \,, $$ counit $$ \varepsilon(c) = 1 \,, \quad \varepsilon(s) = 0 \,, $$ and antipode $$ S(c) = c \,, \quad S(s) = -s \,. $$
We note this in the quotient $\mathbb{R}[c, s] / (c^2 + s^2 - 1)$ our original problem with the antipode doesn’t occur because $$ \sum_{(c)} c_{(1)} S( c_{(2)} ) = c S(c) - s S(s) = c^2 + s^2 = 1 = \varepsilon(c) \,. $$