We consider N × N $N\times N$ tensors for N = 3 , 4 , 5 , 6 $N= 3,4,5,6$ . In the case N = 3 $N=3$ , it is desired to find the three principal invariants i 1 , i 2 , i 3 $i_{1}, i_{2}, i_{3}$ of U ${\mathbf{U}}$ in terms of the three principal invariants I 1 , I 2 , I 3 $I_{1}, I_{2}, I_{3}$ of C = U 2 ${\mathbf{C}}={\mathbf{U}}^{2}$ . Equations connecting the i α $i_{\alpha }$ and I α $I_{\alpha }$ are obtained by taking determinants of the factorisation λ 2 I − C = ( λ I − U ) ( λ I + U ) , $$ \lambda ^{2}{\mathbf{I}}-{\mathbf{C}}= (\lambda {\mathbf{I}}-{\mathbf{U}})( \lambda {\mathbf{I}}+{\mathbf{U}}), $$ and comparing coefficients. On eliminating i 2 $i_{2}$ we obtain a quartic equation with coefficients depending solely on the I α $I_{\alpha }$ whose largest root is i 1 $i_{1}$ . Similarly, we may obtain a quartic equation whose largest root is i 2 $i_{2}$ . For N = 4 $N=4$ we find that i 2 $i_{2}$ is once again the largest root of a quartic equation and so all the i α $i_{\alpha }$ are expressed in terms of the I α $I_{\alpha }$ . Then U ${\mathbf{U}}$ and U − 1 ${\mathbf{U}}^{-1}$ are expressed solely in terms of C ${\mathbf{C}}$ , as for N = 3 $N=3$ . For N = 5 $N= 5$ we find, but do not exhibit, a twentieth degree polynomial of which i 1 $i_{1}$ is the largest root and which has four spurious zeros. We are unable to express the i α $i_{\alpha }$ in terms of the I α $I_{\alpha }$ for N = 5 $N=5$ . Nevertheless, U ${\mathbf{U}}$ and U − 1 ${\mathbf{U}}^{-1}$ are expressed in terms of powers of C ${\mathbf{C}}$ with coefficients now depending on the i α $i_{\alpha }$ . For N = 6 $N=6$ we find, but do not exhibit, a 32 degree polynomial which has largest root i 1 2 $i_{1}^{2}$ . Sixteen of these roots are relevant, which we exhibit, but the other 16 are spurious. U ${\mathbf{U}}$ and U − 1 ${\mathbf{U}}^{-1}$ are expressed in terms of powers of C ${\mathbf{C}}$ . The cases N > 6 $N>6$ are discussed.

中文翻译：

---

${\mathbf{U}} = \mathbf{C}^{1/2}$ 及其在 $\mathbf{C}$ 及其不变量方面的不变量

我们考虑 N × N $N\times N$ 张量 N = 3 , 4 , 5 , 6 $N= 3,4,5,6$ 。在 N = 3 $N=3$ 的情况下，需要找到三个主不变量 i 1 , i 2 , i 3 $i_{1}, i_{2}, i_{3}$ of U ${\ mathbf{U}}$ 的三个主要不变量 I 1 , I 2 , I 3 $I_{1}, I_{2}, I_{3}$ of C = U 2 ${\mathbf{C}} ={\mathbf{U}}^{2}$ 。连接 i α $i_{\alpha }$ 和 I α $I_{\alpha }$ 的方程是通过取因子分解的行列式获得的 λ 2 I − C = ( λ I − U ) ( λ I + U ) , $ $ \lambda ^{2}{\mathbf{I}}-{\mathbf{C}}= (\lambda {\mathbf{I}}-{\mathbf{U}})( \lambda {\mathbf{I }}+{\mathbf{U}})、$$ 和比较系数。消除 i 2 $i_{2}$ 我们得到一个四次方程，其系数完全取决于最大根为 i 1 $i_{1}$ 的 I α $I_{\alpha }$ 。相似地，我们可以得到最大根为 i 2 $i_{2}$ 的四次方程。对于 N = 4 $N=4$ 我们发现 i 2 $i_{2}$ 再次是四次方程的最大根，因此所有 i α $i_{\alpha }$ 都表示为 I α $I_{\alpha }$ 。那么 U ${\mathbf{U}}$ 和 U − 1 ${\mathbf{U}}^{-1}$ 仅用 C ${\mathbf{C}}$ 表示，对于 N = 3 $N=3$ 。对于 N = 5 $N= 5$，我们找到了一个二十次多项式，其中 i 1 $i_{1}$ 是最大的根并且有四个伪零。我们无法用 N = 5 $N=5$ 的 I α $I_{\alpha }$ 来表达 i α $i_{\alpha }$ 。然而，U ${\mathbf{U}}$ 和 U − 1 ${\mathbf{U}}^{-1}$ 用 C ${\mathbf{C}}$ 的幂表示，现在有系数取决于 i α $i_{\alpha }$ 。对于 N = 6 $N=6$ 我们发现，但没有展示，具有最大根 i 1 2 $i_{1}^{2}$ 的 32 度多项式。我们展示了这些根中有 16 个是相关的，但其他 16 个是虚假的。U ${\mathbf{U}}$ 和 U − 1 ${\mathbf{U}}^{-1}$ 以 C ${\mathbf{C}}$ 的幂表示。讨论了 N > 6 $N>6$ 的情况。