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Let $X$ be the set of tetrahedron’s edges and $Y$ the set of its faces. To any function $f$ on $X$ with values in a field $K$, we assign a function $g$ on $Y$ defined as

$g(y)$ = $\sum_{x⊂y} f(x)$

That is, the value of $g$ on a face equals the sum of the values of $f$ on the sides of this face.

The above assignment defines a linear map $φ : F(X, K) → F(Y, K)$, where $F(X, K)$ denotes the space of function from $X$ to $K$ and similarly for $F(Y, K)$

Is The kernel of $φ$ has dimension two if $K$ is the field of rational numbers?

Why the dimension of $F(X,K)$ and $F(Y,K)$ are $6$ and $4$ respectively?

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2 Answers 2

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You can consider the function $1_{l_i}$ that assigns to an edge $x$ the value $0$ if $x\neq l_i$ and $1$ otherwise, for each $i=1, \dots 6$.

If $f$ is a function on $X=\{l_1, \dots , l_6\}$, then you get

$f=\sum_{i=1}^6 f(l_i)\cdot 1_{l_i}$

and so $\{1_{l_1}, \dots , 1_{l_6}\}$ is a base of $F(X, K)$

This is trivial if one observe that $F(X,K)$ corresponds to the free $K-algebra$ generated by the set $X$.

Thus we get the dimension of $F(X,K)$ is equal to $6$, that is exactly the number of tethraedron edges.

An analogous reasoning can be done for $F(Y,K)$, that has dimension $4$ equal to the number of faces.

At this point you can observe $\phi$ is a surjective map. If $g$ is a function on $Y$, then we want to find $f$ such that $\phi(f)=g$, i.e. it is satisfied the following linear system:

$g(F_1)=f(l_1)+f(l_2)+f(l_3)$

$g(F_2)=f(l_3)+f(l_4)+f(l_5)$

$g(F_3)=f(l_4)+f(l_2)+f(l_6)$

$g(F_4)=f(l_1)+f(l_5)+f(l_6)$

and so

$k_1=g(F_1)-g(F_2)+g(F_3)-g(F_4)=2[f(l_2)-f(l_5)]$

that implies, if $K$ has not characteristic $2$, that

$f(l_2)=f(l_5)+\frac{1}{2}k_1$

If we continue the reasoning, we get

$g(F_1)-g(F_2)=f(l_1)+\frac{1}{2}k_1-f(l_4)$

and so

$f(l_1)=f(l_4)+(g(F_1)-g(F_2)-\frac{1}{2}k_1)=f(l_4)+k_2$

Moreover

$f(l_3)=g(F_2)-(f(l_4)+f(l_5))$

$f(l_6)=g(F_4)-k_2-(f(l_4)+f(l_5))$

Thus a general solution of the system will be

$f(l_1):=a+ g(F_1)-g(F_2)-\frac{1}{2}k_1$

$f(l_2):= b+\frac{1}{2}k_1$

$f(l_3):=g(F_2)-(a+b)$

$f(l_4):=a$

$f(l_5):=b$

$f(l_6):=g(F_4)-k_2-(a+b)$

for some fixed $a,b\in K$.

In other words $\phi$ is surjective and so, by nullity rank theorem, we get

$\dim( \ker(\phi))=6-4=2$

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Note that $|X|=6$ and $|Y|=4$ and $F(X,K)=K^6$, while $F(Y,K)=K^4$.

For clarity, let us assume $X=\{(e_1,e_2),(e_1,e_3),(e_1,e_4),(e_2,e_3),(e_2,e_4),(e_3,e_4))\}$ and $Y=\{(e_1,e_2,e_3),(e_1,e_2,e_4),(e_1,e_3,e_4),(e_2,e_3,e_4)\}$ and come to the description of $\varphi$.

It can be described as

$\varphi(\alpha(e_1,e_2,e_3)+\beta(e_1,e_2,e_4)+\gamma(e_1,e_3,e_4)+\delta(e_2,e_3,e_4))\\ =\alpha((e_1,e_2)+(e_2,e_3)+(e_1,e_3))+\beta((e_1,e_2)+\cdots)+\cdots\\ =(\alpha+\beta)(e_1,e_2)+(\alpha+\gamma)(e_1,e_3)+(\beta+\gamma)(e_1,e_4)+(\alpha+\delta)(e_2,e_3)+(\beta+\delta)(e_2,e_4)+(\delta+\gamma)(e_3,e_4)$.

I hope you will be able to conclude from here. Good luck!

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