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It is a bit unclear what you are asking. First, in what way do you interpret $f = 0$ as a projective algebraic curve? It is, as described, not compact. Second, the The thing you call the geometric genus is certainly not the geometric genus; the literal object you wrote is $\infty$ and if you first compactified the curve as I assume you want to do, then it is always $1$. The correct definition of geometric genus would yield $p_g(V) = 0$ in this case, as the map $t \to (t^q, -t^p)$ gives a 1-1 parameterization of your curve, showing that its normalization is $\mathbb{A}^1$.

I suspect the question you mean to ask is the following.

In what way is the $\delta$-invariant of the singularity -- i.e., the local contribution to the arithmetic genus -- related to the dimension of the first homology group of the Milnor fibre?

The answer to this is given by the formula*

$\mathrm{h}_1(f^{-1} (\epsilon) ) =: \mu = 2 \delta + 1 - b$.

I do not know the history of this formula, but it certainly appears in Milnor's book on hypersurface singularities. Here $\delta$ is as above the difference between the arithmetic and geometric genera of a projective curve containing this singularity and smooth elsewhere, and $b$ is the number of analytic local branches, here 1.

*in general, you should intersect $f^{-1}(\epsilon)$ with a small ball, but in the case of $x^p + y^q$ it is unnecessary.

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It is a bit unclear what you are asking. First, in what way do you interpret $f = 0$ as a projective algebraic curve? It is, as described, not compact. Second, the thing you call the geometric genus is certainly not the geometric genus; the literal object you wrote is $\infty$ and if you first compactified the curve as I assume you want to do, then it is always $1$. The correct definition of geometric genus would yield $p_g(V) = 0$ in this case, as the map $t \to (t^q, -t^p)$ gives a 1-1 parameterization of your curve, showing that its normalization is $\mathbb{A}^1$.

I suspect the question you mean to ask is the followsfollowing.

In what way is the $\delta$-invariant of the singularity -- i.e., the local contribution to the arithmetic genus -- related to the dimension of the first homology group of the Milnor fibre. ?

The answer to this is given by the (easy to prove) formula*

$h^0(f^{-1} \mathrm{h}_1(f^{-1} (\epsilon) ) =: \mu = 2 \delta + 1 - b$.

I do not know the history of this formula, but it certainly appears in Milnor's book on hypersurface singularities. Here $\delta$ is as above the difference between the arithmetic and geometric genera of a projective curve containing this singularity and smooth elsewhere, and $b$ is the number of analytic local branches, here 1.

*in general, you should intersect $f^{-1}(\epsilon)$ with a small ball, but in the case of $x^p + y^q$ it is unnecessary.

1

It is a bit unclear what you are asking. First, in what way do you interpret $f = 0$ as a projective algebraic curve? It is, as described, not compact. Second, the thing you call the geometric genus is certainly not the geometric genus; the literal object you wrote is $\infty$ and if you first compactified the curve as I assume you want to do, then it is always $1$. The correct definition of geometric genus would yield $p_g(V) = 0$ in this case, as the map $t \to (t^q, -t^p)$ gives a 1-1 parameterization of your curve, showing that its normalization is $\mathbb{A}^1$.

I suspect the question you mean to ask is the follows. In what way is the $\delta$-invariant of the singularity -- i.e., the local contribution to the arithmetic genus -- related to the dimension of the first homology group of the Milnor fibre. The answer to this is given by the (easy to prove) formula*

$h^0(f^{-1} (\epsilon) ) =: \mu = 2 \delta + 1 - b$.

I do not know the history of this formula, but it certainly appears in Milnor's book on hypersurface singularities. Here $\delta$ is as above the difference between the arithmetic and geometric genera of a projective curve containing this singularity and smooth elsewhere, and $b$ is the number of analytic local branches, here 1.

*in general, you should intersect $f^{-1}(\epsilon)$ with a small ball, but in the case of $x^p + y^q$ it is unnecessary.