The answer is yes.

For the proof notice that $2n$-dimensional sphere $S^{2n}$ is a compactification of $C^{n}$, as well as $S^2$ is a compactification of $C$.

Let $L' \subset S^{2n}$ be the closure of $L$ in the compactification of $C^n$.

Then $L'$ is also a $2$-sphere.

Moreover, $H$ and $g$ extend to the maps between the corresponding compactifications: $H: S^{2n} \to S^{2n}$ and $g: S^{2n} \to S^2$.

Thus the problem is reduced to the following one:

we have a $2$-sphere $L' \subset S^{2n}$ and a composition of maps $F=g\circ H: S^{2n} \to S^{2}$, and we need to prove that the absolute value of the topological degree "topdeg" of $F|_{L'}:S^2\to S^2$ is less or equal than the algebraic degree "algdeg" of $g$.

1) Suppose $H:S^{2n} \to S^{2n}$ preserves orientation.

Then it is isotopic to the identity map $id(S^2)$, whence since the topological degree is a homotopy invariant, we obtain that

$r = topdeg(F|_{L'}) = topdeg(g \circ id|_{L'}) = topdeg(g|_{L'}) \leq algdeg(g) = k$.

2) If $H$ reverses orientation, then it is isotopic to a reflection with respect to an equator $S^{2n-1}$, which corresponds to a $(2n-1)$-dimensional real subspace of $C^{n} = R^{2n}$. In other words $H$ is isotopic to a conjugation, say of the first complex coordinate:

$Q:C^{n} \to C^{n}$, $q(z_1,z_2,\ldots,z_n) = (\bar{z}_1,z_2,\ldots,z_n)$.

Now we have two cases.

2a) Suppose $n>1$.

Then we can assume that $L$ is contained in a subspace $\{ z_1 = 0 \}$, so $Q|_L = id_{L}$, whence again

$r = topdeg(F|_{L'}) = topdeg(g \circ Q|_{L'}) = topdeg(g \circ id|_{L'})  = topdeg(g|_{L'}) \leq algdeg(g) = k$.

2b) Let $n=1$.

Then $H:S^2\to S^2$ is a reversing orientation homeomorphism of the $2$-sphere, so $topdeg(H) = topdeg(Q) = -1$. Whence

$r = topdeg(F|_{L'}) = topdeg(g \circ Q) = - topdeg(g) = -algdeg(g) = -k$.

Thus in all the cases $|topdeg(F|_{L})| \leq algdeg(g)$.

Hello,

Thank you for your attention.

I have questions:

Why $topdeg (g\circ id|_{L '}) \ leq topdeg (g|_{L'}) $?

I think in cases 1) and 2a) we have r = k. Is that correct?

Again thank you.

Best regards,

Edson Sampaio

Hi Edson. You are right, in 1) and 2) instead of

$ r = ... = topdeg(g \circ id|_{L'}) \leq topdeg(g|_{L'}) = algdeg(g) = k$

there must be

$ r = ... = topdeg(g \circ id|_{L'}) = topdeg(g|_{L'}) \leq algdeg(g) = k$

I corrected the answer.

On the other hand $r$ can be not equal to $k$, that is it may happens that

$r = topdeg(g|_{L'}) < algdeg(g) = k$.

It seems that in this case we always have that $r=0$, but I can not produce at least just now an example when $0

There is a problem! How extend $g$ to the map between the corresponding compactifications, $g: S^{2n} \to S^2$? Observe that if $x\in g^{-1}(0)\setminus \{0\}$, then $tx\in  g^{-1}(0)$ for all $t\in C$, in particular, $\lim\limits_{k\to \infty} g(kx)=0$, i.e., $g$ is not a proper map, if $n>1$, right?

@Edson Sampaio. You are right, a homogeneous polynomial need not be proper if n > 1. in fact, what do you  mean  precisely with the topological degree of g: L --> C ?The usual way to do this is to  have some topologically invariant abelian group of rank1 and a map between them functorially defined.

The most obvious group here is Cohomology with compact support H_c^i . We have a canonical isomorphism H_c^2(L) = H_c^2(L) = Z given by complex orientation, but in the absense of properness of H there is no obvious map. g_*H^i_c(L) --> H^i_c(C) .  There is however a non obvious map g_! = H_!\phi_! H^i_c(L) -->H^i_c(C). The map \phi_! is essentially \pm the identity because \phi is an isomorphism, and  if I am not terribly mistaken, the map H_! is constant if we deform H: C^n --> C , to be transversal , e.g to H = K L with K: (z_1,..., z_n) --> z_1 and L: z_1 --> z_1^k. In this case K_! is the Gysin map K_!  Z = H^2n_c(C^n)  --> H^2_c(C) = Z,   which after identification sends1 to 1, and L_! H^2_c(C) --> H^2_c(C) is multiplication with k.

So interpreted this way we would indeed have deg(g) = \pm k.

@Rogier Brussee. Thanks for your response. But I did not understand, I think you changed the notation. Can you explain a bit more your idea?