[proofplan] The smooth proper family over the punctured disk defines a rational cohomology local system, and the chosen positive loop around the origin gives the operators $T_k$. The deep input is the Kähler [monodromy theorem](/theorems/3371): for a proper Kähler degeneration, the Gauss-Manin connection has regular singular behaviour and its cohomological monodromy is quasi-unipotent. Once the eigenvalues are roots of unity, a single finite base change kills the finite-order semisimple parts in all nonzero cohomological degrees, leaving unipotent monodromy. The logarithm is then a finite polynomial in the nilpotent operator $T_k^m-I$, and this polynomial is itself nilpotent because it is divisible by $T_k^m-I$. [/proofplan] [step:Construct the cohomological monodromy local system over the punctured disk] Set \begin{align*} U:=f^{-1}(\Delta^*). \end{align*} The restricted morphism \begin{align*} f_U:=f|_U:U\to\Delta^* \end{align*} is smooth and proper by hypothesis. Since $f$ is a proper holomorphic Kähler morphism, each fibre $X_t=f^{-1}(t)$ is a compact Kähler manifold. The smooth proper morphism $f_U$ therefore gives, for each integer $k\ge 0$, a finite-dimensional rational local system \begin{align*} \mathbb V_k:=R^k(f_U)_*\mathbb Q \end{align*} on $\Delta^*$, whose fibre at $t$ is naturally $H^k(X_t,\mathbb Q)$. The positive generator of $\pi_1(\Delta^*,t)$ is represented by the positively oriented loop once around $0$. Parallel transport in the local system $\mathbb V_k$ along this loop gives the automorphism \begin{align*} T_k:H^k(X_t,\mathbb Q)\to H^k(X_t,\mathbb Q). \end{align*} This is exactly the monodromy operator appearing in the statement. [guided] Let \begin{align*} U:=f^{-1}(\Delta^*). \end{align*} We use the notation \begin{align*} f_U:=f|_U:U\to\Delta^* \end{align*} for the restriction of $f$ to the punctured disk. This avoids the misleading notation $f^*$, which normally denotes pullback. By hypothesis, $f_U$ is smooth, and since $f$ is proper, the restriction $f_U$ is also proper. The phrase “Kähler morphism” is used here in the standard sense that supplies relative Kähler geometry on the fibres; in particular, the fibres \begin{align*} X_t=f^{-1}(t) \end{align*} are compact Kähler manifolds. For every integer $k\ge 0$, define \begin{align*} \mathbb V_k:=R^k(f_U)_*\mathbb Q. \end{align*} This is the rational cohomology local system of the smooth proper family over $\Delta^*$. Its fibre at the chosen base point $t\in\Delta^*$ is \begin{align*} (\mathbb V_k)_t=H^k(X_t,\mathbb Q). \end{align*} A local system has parallel transport along paths. The fundamental group $\pi_1(\Delta^*,t)$ is generated by the positive loop going once around the deleted origin. Transporting the fibre $H^k(X_t,\mathbb Q)$ along that loop gives a rational linear automorphism \begin{align*} T_k:H^k(X_t,\mathbb Q)\to H^k(X_t,\mathbb Q). \end{align*} This construction is the precise meaning of the cohomological monodromy operator in the theorem. [/guided] [/step] [step:Apply the Kähler monodromy theorem to obtain quasi-unipotence] We invoke the monodromy theorem for proper Kähler degenerations: for a proper holomorphic Kähler morphism over the disk whose restriction over the punctured disk is smooth, the monodromy acting on each rational cohomology group of a smooth fibre is quasi-unipotent. Equivalently, every eigenvalue of each operator $T_k$ is a root of unity. The hypotheses of this theorem match the present situation: $f:\mathcal X\to\Delta$ is proper and Kähler, and $f_U:U\to\Delta^*$ is smooth. Hence, for every $k\ge 0$, every eigenvalue of \begin{align*} T_k:H^k(X_t,\mathbb Q)\to H^k(X_t,\mathbb Q) \end{align*} is a root of unity. This proves the first assertion. Here the cited result is the standard Kähler monodromy theorem, equivalently the quasi-unipotence theorem obtained from regular singularity of the Gauss-Manin connection and the rational cohomology lattice. This result is not yet available in the wiki as a resolved theorem citation. [/step] [step:Choose one base-change exponent that works in every cohomological degree] Let $n$ be the complex dimension of the compact fibre $X_t$. Since $X_t$ is a compact real manifold of real dimension $2n$, one has \begin{align*} H^k(X_t,\mathbb Q)=0 \end{align*} for every $k>2n$. Thus only the finitely many degrees $0\le k\le 2n$ need to be considered. For each $k$ with $0\le k\le 2n$, let $S_k\subset\mathbb C$ be the finite set of eigenvalues of $T_k$ on the complex [vector space](/page/Vector%20Space) \begin{align*} H^k(X_t,\mathbb Q)\otimes_{\mathbb Q}\mathbb C. \end{align*} By the preceding step, every $\lambda\in S_k$ is a root of unity. Let $m\ge 1$ be a common multiple of the orders of all roots of unity in the finite set \begin{align*} \bigcup_{k=0}^{2n}S_k. \end{align*} Then, for every $k\ge 0$ and every eigenvalue $\lambda$ of $T_k$, one has \begin{align*} \lambda^m=1. \end{align*} Therefore every eigenvalue of $T_k^m$ is equal to $1$. [/step] [step:Identify the monodromy after the finite base change] Consider the finite holomorphic map \begin{align*} \beta:\Delta&\to\Delta \end{align*} \begin{align*} s&\mapsto s^m. \end{align*} Over the punctured disk it restricts to \begin{align*} \beta^*:\Delta^*&\to\Delta^* \end{align*} \begin{align*} s&\mapsto s^m. \end{align*} Let \begin{align*} \mathcal X_\beta:=\mathcal X\times_{\Delta,\beta}\Delta \end{align*} be the base-changed total space, and let \begin{align*} f_\beta:\mathcal X_\beta\to\Delta \end{align*} be the induced morphism. Over $\Delta^*$, the positive generator of $\pi_1(\Delta^*,s)$ maps under $\beta^*$ to the $m$-fold positive generator of $\pi_1(\Delta^*,\beta(s))$. Hence parallel transport in the pulled-back local system around the positive generator is the $m$-fold iterate of the original parallel transport. Therefore the monodromy of the base-changed family on $H^k(X_t,\mathbb Q)$ is \begin{align*} T_k^m. \end{align*} Since every eigenvalue of $T_k^m$ is $1$, the [characteristic polynomial](/page/Characteristic%20Polynomial) of $T_k^m$ on the finite-dimensional complex vector space $H^k(X_t,\mathbb Q)\otimes_{\mathbb Q}\mathbb C$ is a power of $x-1$. By the [Cayley-Hamilton theorem](/theorems/865) applied to $T_k^m-I$, there exists an integer $r_k\ge 1$ such that \begin{align*} (T_k^m-I)^{r_k}=0. \end{align*} Thus $T_k^m$ is unipotent for every $k\ge 0$. [guided] The base change is the map \begin{align*} \beta:\Delta&\to\Delta \end{align*} \begin{align*} s&\mapsto s^m. \end{align*} Over the punctured disk this is an $m$-fold covering map. Let \begin{align*} \mathcal X_\beta:=\mathcal X\times_{\Delta,\beta}\Delta \end{align*} be the fibre product, and let \begin{align*} f_\beta:\mathcal X_\beta\to\Delta \end{align*} be the induced morphism. The cohomology local system of $f_\beta$ over $\Delta^*$ is the pullback of the original local system $\mathbb V_k$ along $\beta^*$. We now compute the monodromy of this pulled-back local system. A positive generator of $\pi_1(\Delta^*,s)$ is a loop in the $s$-disk going once counterclockwise around $0$. Applying $s\mapsto s^m$ sends this loop to a loop in the original punctured disk that winds $m$ times counterclockwise around $0$. Parallel transport around an $m$-fold loop is the composition of the original monodromy with itself $m$ times. Therefore the new monodromy on the fibre cohomology is \begin{align*} T_k^m:H^k(X_t,\mathbb Q)\to H^k(X_t,\mathbb Q). \end{align*} It remains to justify that this operator is unipotent. From the choice of $m$, every eigenvalue $\lambda$ of $T_k$ satisfies $\lambda^m=1$. Hence every eigenvalue of $T_k^m$ is $1$. Equivalently, every eigenvalue of \begin{align*} T_k^m-I \end{align*} is $0$. On the finite-dimensional complex vector space \begin{align*} H^k(X_t,\mathbb Q)\otimes_{\mathbb Q}\mathbb C, \end{align*} the characteristic polynomial of $T_k^m-I$ is therefore a power of $x$. By the [Cayley-Hamilton theorem](/theorems/923), the operator $T_k^m-I$ satisfies its characteristic polynomial, so some positive power of $T_k^m-I$ is zero. Thus there exists $r_k\ge 1$ with \begin{align*} (T_k^m-I)^{r_k}=0. \end{align*} This is precisely the assertion that $T_k^m$ is unipotent. [/guided] [/step] [step:Define the logarithm as a finite polynomial and prove it is nilpotent] Fix $k\ge 0$, and define \begin{align*} A_k:=T_k^m-I. \end{align*} From the previous step, choose $r_k\ge 1$ such that \begin{align*} A_k^{r_k}=0. \end{align*} The logarithmic series of $I+A_k=T_k^m$ terminates because $A_k$ is nilpotent, so define \begin{align*} N_k:=\log(T_k^m):=\sum_{j=1}^{r_k-1}(-1)^{j+1}\frac{A_k^j}{j}. \end{align*} This is a well-defined endomorphism of $H^k(X_t,\mathbb Q)$. Define the polynomial \begin{align*} q_k(x):=\sum_{j=1}^{r_k-1}(-1)^{j+1}\frac{x^{j-1}}{j}\in\mathbb Q[x]. \end{align*} Then \begin{align*} N_k=A_k q_k(A_k). \end{align*} Since $A_k$ commutes with every polynomial in $A_k$, we have \begin{align*} N_k^{r_k}=A_k^{r_k}q_k(A_k)^{r_k}=0. \end{align*} Thus $N_k$ is nilpotent. This proves all assertions of the theorem. [/step]